RS Aggarwal Solutions for Class 10, maths Chapter 8

Question 1:

A point P is at a distance of 29 cm from the centre of a circle of radius 20 cm. Find the length of the tangent drawn from P to the circle. [CBSE 2017]

Answer:

Consider the figure.
We know that the tangent is perpendicular to the radius of a circle.
So, OPB is a right angled triangle, with $∠ OBP = 90 °$
By using pythagoras theorem in $△ OPB$ , we get
$\Rightarrow {OB}^{2} + {PB}^{2} = {OP}^{2} \Rightarrow (20)^{2} + {PB}^{2} = (29)^{2} \Rightarrow 400 + {PB}^{2} = 841 \Rightarrow {PB}^{2} = 841 - 400 = 441 \Rightarrow PB = \sqrt{441} = 21$
So, length of the tangent from point P is 21 cm.

Question 2:

A point P is 25 cm away from the centre of a circle and the length of tangent drawn from P to the circle is 24 cm. Find the radius of the circle.

Answer:

$Draw a circle and let P be a point such that OP = 25 cm . Let TP be the tangent, so that TP = 24 cm Join OT,where OT is radius . Now, tangent drawn from an external point is perpendicular to the radius at the point of contact . ∴ OT ⊥ PT In the right △ OTP, we have: {OP}^{2} = {OT}^{2} + {TP}^{2} [By Pythagoras’ theorem:] {OT}^{2} = \sqrt{{OP}^{2} - {TP}^{2}} = \sqrt{25^{2} - 24^{2}} = \sqrt{625 - 576} = \sqrt{49} =7 cm ∴ The length of the radius is 7 cm .$

Question 3:

Two concentric circles are of radii 6.5 cm and 2.5 cm. Find the length of the chord of the larger circle which touches the smaller circle

Answer:

We know that the radius and tangent are perperpendular at their point of contact
In right triangle AOP
AO² = OP² + PA²
⇒ (6.5)² = (2.5)² + PA²
⇒ PA² = 36
⇒ PA = 6 cm
Since, the perpendicular drawn from the centre bisect the chord.
∴ PA = PB = 6 cm
Now, AB = AP + PB = 6 + 6 = 12 cm
Hence, the length of the chord of the larger circle is 12 cm.

Question 4:

In the given figure, a circle inscribed in a triangle ABC, touches the sides AB, BC and AC at point D, E and F respectively. If AB = 12 cm, BC = 8 cm and AC = 10 cm, find the lengths of AD, BE and CF [CBSE 2013]

Answer:

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AD = AF, BD = BE and CE = CF
Now, AD + BD = 12 cm          …..(1)
AF + FC = 10 cm
⇒ AD + FC = 10 cm                    …..(2)
BE + EC = 8 cm
⇒ BD + FC = 8 cm                   …..(3)
Adding all these we get
AD + BD + AD + FC + BD + FC = 30
⇒2(AD + BD + FC) = 30
⇒AD + BD + FC = 15 cm           …..(4)
Solving (1) and (4), we get
FC = 3 cm
Solving (2) and (4), we get
BD = 5 cm
Solving (3) and (4), we get
and AD = 7 cm
∴ AD = AF = 7 cm, BD = BE = 5 cm and CE = CF = 3 cm

Question 5:

In the given figure, PA and PB are the tangents to a circle with centre O. Show that the points A, O, B, P are concyclic.

Answer:

$Here, O A = O B A nd O A ⊥ A P, O A ⊥ B P (S ince tangents drawn from an external point are perpendicular to the radius at the point of contact) ∴ ∠ O A P = 90^{0}, ∠ O B P = 90^{0} ∴ ∠ O A P + ∠ O B P = 90^{0} + 90^{0} = 180^{0} ∴ ∠ A O B + ∠ A P B = 180^{0} (S ince, ∠ O A P + ∠ O B P + ∠ A O B + ∠ A P B = 360^{0}) S um of opposite angle of a quadrilateral is 180 ° . Hence, A, O, B and P are concyclic .$

Question 6:

In the given figure, the chord AB of the larger of the two concentric circles, with centre O, touches the smaller circle at C. Prove that AC = CB.

Answer:

Construction: Join OA, OC and OB

We know that the radius and tangent are perperpendular at their point of contact
∴ ∠OCA = ∠OCB = 90^∘
Now, In △OCA and △OCB
∠OCA = ∠OCB = 90^∘
OA = OB (Radii of the larger circle)
OC = OC (Common)
By RHS congruency
△OCA ≅ △OCB
∴ CA = CB

Question 7:

From an external point P, tangents PA and PB are drawn to a circle with centre O. If CD is the tangent to the circle at a point E and PA = 14 cm, find the perimeter of ΔPCD.

Answer:

$Given, PA and PB are the tangents to a circle with centre O and CD is a tangent at E and PA = 14 cm . Tangents drawn from an external point are equal. ∴ PA = PB,CA = CE and DB = DE Perimeter of △ PCD = PC + CD + PD = (PA - CA) + (CE + DE) + (PB - DB) = (PA - C E) + (CE + DE) + (PB - DE) = (PA + PB) =2PA (∵ PA=PB) = (2 \times 14) cm =28 cm ∴ Perimeter of △ PCD = 28 cm .$

Question 8:

A circle is inscribed in ΔABC, touching AB, BC and AC at P, Q and R, respectively. If AB = 10 cm, AR = 7 cm and CR = 5 cm, find the length of BC.

Answer:

$Given, a circle inscribed in triangle ABC, such that the circle touches the sides of the triangle at P,Q,R . Tangents drawn to a circle from an external point are equal . ∴ AP = AR = 7 cm, CQ = CR = 5 cm . Now, BP = (AB - AP) = (10 - 7) = 3 cm ∴ BP = BQ = 3 cm ∴ BC = (BQ + QC) = > B C = 3 + 5 = > B C = 8 ∴ The length of BC is 8 cm .$

Question 9:

In the given figure, a circle touches all the four sides of a quadrilateral ABCD whose three sides are AB = 6 cm, BC = 7 cm and CD = 4 cm. Find AD.

Answer:

$Let the circle touch the sides of the quadrilateral AB,BC,CD and DA at P,Q,R and S respectively . Given, A B = 6 cm, B C = 7 cm and C D = 4 cm . Tangents drawn from an external point are equal. ∴ A P = A S, B P = B Q, C R = C Q and D R = D S \begin{array}{l} Now, A B + C D = (A P + B P) + (C R + D R) \\ = > A B + C D = (A S + B Q) + (C Q + D S) \\ = > A B + C D = (A S + D S) + (B Q + C Q) \\ = > A B + C D = A D + B C \\ = > A D = (A B + C D) - B C \\ = > A D = (6 + 4) - 7 \\ = > A D = 3 cm . \end{array} ∴ The length of A D is 3 cm .$

Question 10:

In the given figure, an isosceles triangle ABC with AB = AC, circumscribes a circle. Prove that the point of contact P bisects the base BC
[CBSE 2012]

Answer:

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AR = AQ, BR = BP and CP = CQ
Now, AB = AC
⇒ AR + RB = AQ + QC
⇒ AR + RB = AR + QC
⇒ RB = QC
⇒ BP = CP
Hence, P bisects BC at P.

Question 11:

In the given figure, O is the centre of two concentric circles of radii 4 cm and 6 cm respectively. PA and PB are tangents to the outer and inner circles, respectively. If PA = 10 cm, find the length of PB up to one decimal place.

Answer:

$Given, O is the centre of two concentric circles of radii O A = 6 cm and O B = 4 cm . P A and P B are the two tangents to the outer and inner circles respectively and P A = 10 cm . Now, tangent drawn from an external point is perpendicular to the radius at the point of contact . ∴ ∠ O A P = ∠ O B P = 90^{0} ∴ From right-angled △ O A P, O P^{2} = O A^{2} + P A^{2} \begin{array}{l} = > O P = \sqrt{O A^{2} + P A^{2}} \\ = > O P = \sqrt{6^{2} + 10^{2}} \\ = > O P = \sqrt{136} cm . \end{array} ∴ From right-angled △ O A P, O P^{2} = O B^{2} + P B^{2} \begin{array}{l} = > P B = \sqrt{O P^{2} - O B^{2}} \\ = > P B = \sqrt{136 - 16} \\ = > P B = \sqrt{120} cm \\ => P B = 10.9 cm . \end{array} ∴ The length of P B is 10.9 cm .$

Question 12:

In the given figure, a triangle ABC is drawn to circumscribe a circle of radius 3 cm such that the segments BD and DC into which BC is divided by the point of contact D are, of lengths 6 cm and 9 cm respectively. If the area of △ABC = 54 cm² then find the lengths of sides of AB and AC. [CBSE 2011, ’15]

Answer:

Construction: Join OA, OB, OC, OE ⊥ AB at E and OF ⊥ AC at F

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AE = AF, BD = BE = 6 cm and CD = CF = 9 cm
Now,
$Area (△ ABC) = Area (△ BOC) + Area (△ AOB) + Area (△ AOC) \Rightarrow 54 = \frac{1}{2} \times BC \times OD + \frac{1}{2} \times AB \times OE + \frac{1}{2} \times AC \times OF \Rightarrow 108 = 15 \times 3 + (6 + x) \times 3 + (9 + x) \times 3 \Rightarrow 36 = 15 + 6 + x + 9 + x$

$\Rightarrow 36 = 30 + 2 x \Rightarrow 2 x = 6 \Rightarrow x = 3 cm$
∴ AB = 6 + 3 = 9cm and AC = 9 + 3 = 12 cm

Question 13:

PQ is a chord of length 4.8 cm of a circle of radius 3 cm. The tangent at P and Q intersect at a point T as shown in the figure. Find the length of TP [CBSE 2013C]

Answer:

Let TR = y and TP = x
We know that the perpendicular drawn from the centre to the chord bisects it.
∴ PR = RQ
Now, PR + RQ = 4.8
⇒ PR + PR = 4.8
⇒ PR = 2.4
Now, in right triangle POR
By Using Pyhthagoras theorem, we have
PO² = OR² + PR²
⇒ 3² = OR² + (2.4)²
⇒ OR² = 3.24
⇒ OR = 1.8
Now, in right triangle TPR
By Using Pyhthagoras theorem, we have
TP² = TR² + PR²
⇒ x² = y² + (2.4)²
⇒ x² = y² + 5.76 …..(1)
Again, in right triangle TPQ
By Using Pyhthagoras theorem, we have
TO² = TP² + PO²
⇒ (y + 1.8)² = x² + 3²
⇒ y² + 3.6y + 3.24 = x² + 9
⇒ y² + 3.6y = x² + 5.76 …..(2)
Solving (1) and (2), we get
x = 4 cm and y = 3.2 cm
∴ TP = 4 cm

Question 14:

Prove that the line joining the points of contact of two parallel tangents of a circle passes through its centre. [CBSE 2014]

Answer:

Suppose CD and AB are two parallel tangents of a circle with centre O
Construction: Draw a line parallel to CD passing through O i.e, OP
We know that the radius and tangent are perperpendular at their point of contact.
∠OQC = ∠ORA = 90^∘
Now, ∠OQC + ∠POQ = 180^∘ (co-interior angles)
⇒ ∠POQ = 180^∘ − 90^∘ = 90^∘
Similarly, Now, ∠ORA + ∠POR = 180^∘ (co-interior angles)
⇒ ∠POR = 180^∘ − 90^∘ = 90^∘
Now, ∠POR + ∠POQ = 90^∘ + 90^∘ = 180^∘
Since, ∠POR and ∠POQ are linear pair angles whose sum is 180^∘
Hence, QR is a straight line passing through centre O.

Question 15:

In the given figure, a circle with centre O, is inscribed in a quadrilateral ABCD such that it touches the side BC, AB, AD and CD at points P, Q, R and S respectively. If AB = 29 cm, AD = 23 cm, ∠B = 90^∘ and DS = 5 cm then find the radius of the circle. [CBSE 2008, 13]

Answer:

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
DS = DR, AR = AQ
Now, AD = 23 cm
⇒ AR + RD = 23
⇒ AR = 23 − RD
⇒ AR = 23 − 5 [∵ DS = DR = 5]
⇒ AR = 18 cm
Again, AB = 29 cm
⇒ AQ + QB = 29
⇒ QB = 29 − AQ
⇒ QB = 29 − 18 [∵ AR = AQ = 18]
⇒ QB = 11 cm
Since all the angles are in a quadrilateral BQOP are right angles and OP = BQ.
Hence, BQOP is a square.
We know that all the sides of square are equal.
Therefore, BQ = PO = 11 cm
Hence, the radius of the circle is 11 cm.

Question 16:

In the given figure, O is the centre of the circle and TP is the tangent to the circle from an external point T. If ∠PBT = 30^∘ , prove that
BA : AT = 2 : 1 [CBSE 2015]

Answer:

AB is the chord passing through the centre
So, AB is the diameter
Since, angle in a semi circle is a right angle
∴∠APB = 90^∘
By using alternate segment theorem
We have ∠APB = ∠PAT = 30^∘
Now, in △APB
∠BAP + ∠APB + ∠BAP = 180^∘ (Angle sum property of triangle)
⇒ ∠BAP = 180^∘ − 90^∘ − 30^∘ = 60^∘
Now, ∠BAP = ∠APT + ∠PTA (Exterior angle property)
⇒ 60^∘ = 30^∘ + ∠PTA
⇒ ∠PTA = 60^∘ − 30^∘ = 30^∘
We know that sides opposite to equal angles are equal.
∴ AP = AT
In right triangle ABP
$\sin ∠ ABP = \frac{AP}{BA} \Rightarrow \sin 30 ° = \frac{AT}{BA} \Rightarrow \frac{1}{2} = \frac{AT}{BA}$
∴ BA : AT = 2 : 1

Question 1:

In the adjoining figure, a circle touches all the four sides of a quadrilateral ABCD whose sides are AB = 6 cm, BC = 9 cm and CD = 8 cm. Find the length of AD [CBSE 2011]

Answer:

We know that when a quadrilateral circumscribes a circle then sum of opposites sides is equal to the sum of other opposite sides.
∴ AB + CD = AD + BC
⇒6 + 8 = AD + 9
⇒ AD = 5 cm

Question 2:

In the given figure, PA and PB are two tangents to the circle with centre O. If ∠APB = 50^∘ then what is the measure of ∠OAB is [CBSE 2015]

Answer:

Construction: Join OB

We know that the radius and tangent are perperpendular at their point of contact
∵∠OBP = ∠OAP = 90^∘
Now, In quadrilateral AOBP
∠AOB + ∠OBP + ∠APB + ∠OAP = 360^∘            [Angle sum property of a quadrilateral]
⇒ ∠AOB + 90^∘ + 50^∘+ 90^∘ = 360^∘
⇒ 230^∘+ ∠BOC = 360^∘
⇒ ∠AOB = 130^∘
Now, In isoceles triangle AOB
∠AOB + ∠OAB + ∠OBA = 180^∘            [Angle sum property of a triangle]
⇒ 130^∘ + 2∠OAB = 180⁰                          [∵∠OAB = ∠OBA ]
⇒ ∠OAB = 25^∘

Question 3:

In the given figure, O is the centre of a circle PT and PQ are tangents to the circle from an external point P. If ∠TPQ = 70^∘ then ∠TRQ
[CBSE 2015]

Answer:

Construction: Join OQ and OT

We know that the radius and tangent are perperpendular at their point of contact
∵∠OTP = ∠OQP = 90^∘
Now, In quadrilateral OQPT
∠QOT + ∠OTP + ∠OQP + ∠TPQ = 360^∘ [Angle sum property of a quadrilateral]
⇒ ∠QOT + 90^∘ + 90^∘ + 70^∘ = 360^∘
⇒ 250^∘ + ∠QOT = 360^∘
⇒ ∠QOT = 110^∘
We know that the angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.
$∴ ∠ TRQ = \frac{1}{2} (∠ QOT) = 55 °$

Question 4:

In the given figure, common tangents AB and CD to the two circle with centres O₁ and O₂ intersect at E. Prove that AB = CD [CBSE 2014]

Answer:

We know that tangent segments to a circle from the same external point are congruent.
So, we have
EA = EC for the circle having centre O₁
and
ED = EB for the circle having centre O₁
Now, Adding ED on both sides in EA = EC, we get
EA + ED = EC + ED
⇒EA + EB = EC + ED
⇒AB = CD

Question 5:

If PT is a tangent to a circle with centre O and PQ is a chord of the circle such that ∠QPT = 70^∘ then find the measure of ∠POQ

Answer:

We know that the radius and tangent are perperpendular at their point of contact.
∴∠OPT = 90^∘
Now, ∠OPQ = ∠OPT − ∠TPQ = 90^∘ − 70^∘ = 20^∘
Since, OP = OQ as both are radius
∴∠OPQ = ∠OQP = 20^∘ (Angles opposite to equal sides are equal)
Now, In isosceles △POQ
∠POQ + ∠OPQ + ∠OQP = 180^∘ (Angle sum property of a triangle)
⇒ ∠POQ = 180^∘ − 20^∘ − 20^∘ = 140^∘

Question 6:

In the given figure, a triangle ABC is drawn to circumscribe a circle of radius 2 cm such that the segments BD and DC into which BC is divided by the point of contact D are, of lengths 4 cm and 3 cm respectively. If the area of △ABC = 21 cm² then find the lengths of sides of AB and AC. [CBSE 2011]

Answer:

Construction: Join OA, OB, OC, OE ⊥ AB at E and OF ⊥ AC at F

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AE = AF, BD = BE = 4 cm and CD = CF = 3 cm
Now,
$Area (△ ABC) = Area (△ BOC) + Area (△ AOB) + Area (△ AOC) \Rightarrow 21 = \frac{1}{2} \times BC \times OD + \frac{1}{2} \times AB \times OE + \frac{1}{2} \times AC \times OF \Rightarrow 42 = 7 \times 2 + (4 + x) \times 2 + (3 + x) \times 2 \Rightarrow 21 = 7 + 4 + x + 3 + x$

$\Rightarrow 21 = 14 + 2 x \Rightarrow 2 x = 7 \Rightarrow x = 3.5 cm$
∴ AB = 4 + 3.5 = 7.5 cm and AC = 3 + 3.5 = 6.5 cm

Question 7:

Two concentric circles are of radii 5 cm and 3 cm, respectively. Find the length of the chord of the larger circle that touches the smaller circle.

Answer:

Given: Two circles have the same centre O and AB is a chord of the larger circle touching the
smaller circle at C; also, OA=5 cm and OC=3 cm.
$In ∆ O A C, O A^{2} = O C^{2} + A C^{2} ∴ A C^{2} = O A^{2} - O C^{2} \Rightarrow A C^{2} = 5^{2} - 3^{2} \Rightarrow A C^{2} = 25 - 9 \Rightarrow A C^{2} = 16 \Rightarrow A C = 4 cm ∴ A B = 2 A C (since perpendicular drawn from the centre of the circle bisects the chord) ∴ A B = 2 \times 4 = 8 cm$
The length of the chord of the larger circle is 8 cm.

Question 8:

Prove that the perpendicular at the point of contact of the tangent to a circle passes through the centre.

Answer:

Let AB be the tangent to the circle at point P with centre O.
To prove: PQ passes through the point O.
Construction: Join OP.
Through O, draw a straight line CD parallel to the tangent AB.
Proof: Suppose that PQ doesn’t passes through point O.
PQ intersect CD at R and also intersect AB at P.
AS, CD ∥ AB, PQ is the line of intersection,
∠ORP = ∠RPA (Alternate interior angles)
but also,
∠RPA = 90^∘ (OP ⊥ AB)
⇒ ∠ORP = 90^∘
∠ROP + ∠OPA = 180^∘ (Co interior angles)
⇒∠ROP + 90^∘ = 180^∘
⇒∠ROP = 90^∘
Thus, the ΔORP has 2 right angles i.e. ∠ORP and ∠ROP which is not possible.
Hence, our supposition is wrong.
∴ PQ passes through the point O.

Question 9:

In the given figure, two tangents RQ and RP are drawn from an external point R to the circle with centre O. If ∠PRQ = 120^∘ then prove that OR = PR + RQ

Answer:

Construction: Join PO and OQ
In △POR and △QOR
OP = OQ   (Radii)
RP = RQ    (Tangents from the external point are congruent)
OR = OR    (Common)
By SSS congruency, △POR ≅ △QOR
∠PRO = ∠QRO    (C.P.C.T)
Now, ∠PRO + ∠QRO = ∠PRQ
⇒ 2∠PRO = 120^∘
⇒ ∠PRO = 60^∘
Now, In △POR
$\cos 60^{°} = \frac{PR}{OR} \Rightarrow \frac{1}{2} = \frac{PR}{OR} \Rightarrow OR = 2 PR \Rightarrow OR = PR + PR \Rightarrow OR = PR + RQ$

Question 10:

In the given figure, a circle inscribed in a triangle ABC, touches the sides AB, BC and AC at point D, E and F respectively. If AB = 14 cm, BC = 8 cm and AC = 12 cm. Find the lengths of AD, BE and CF [CBSE 2013]

Answer:

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AD = AF, BD = BE and CE = CF
Now, AD + BD = 14 cm          …..(1)
AF + FC = 12 cm
⇒ AD + FC = 12 cm                    …..(2)
BE + EC = 8 cm
⇒ BD + FC = 8 cm                   …..(3)
Adding all these we get
AD + BD + AD + FC + BD + FC = 34
⇒2(AD + BD + FC) = 34
⇒AD + BD + FC = 17 cm           …..(4)
Solving (1) and (4), we get
FC = 3 cm
Solving (2) and (4), we get
BD = 5 cm = BE
Solving (3) and (4), we get
and AD = 9 cm

Question 11:

In the given figure, O is the centre of the circle. PA and PB are tangents. Show that AOBP is a cyclic quadrilateral [CBSE 2014]

Answer:

We know that the radius and tangent are perpendicular at their point of contact
∵∠OBP = ∠OAP = 90^∘
Now, In quadrilateral AOBP
∠APB + ∠AOB + ∠OBP + ∠OAP = 360^∘ [Angle sum property of a quadrilateral]
⇒ ∠APB + ∠AOB + 90^∘ + 90^∘= 360^∘
⇒ ∠APB + ∠AOB = 180^∘
Also, ∠OBP + ∠OAP = 180^∘
Since, the sum of the opposite angles of the quadrilateral is 180^∘
Hence, AOBP is a cyclic quadrilateral.

Question 12:

In two concentric circles, a chord of length 8 cm of the larger circle touches the smaller circle. If the radius of the larger circle is 5 cm then Find the radius of the smaller circle. [CBSE 2013C]

Answer:

We know that the radius and tangent are perperpendular at their point of contact
Since, the perpendicular drawn from the centre bisect the chord.
∴ AP = PB = $\frac{AB}{2}$ = 4 cm
In right triangle AOP
AO² = OP² + PA²
⇒ 5² = OP² + 4²
⇒ OP² = 9
⇒ OP = 3 cm
Hence, the radius of the smaller circle is 3 cm.

Question 13:

In the given figure, PQ is a chord of a circle with centre O and PT is a tangent. If ∠QPT = 60^∘ , find ∠PRQ

Answer:

We know that the radius and tangent are perperpendular at their point of contact.
∴∠OPT = 90^∘
Now, ∠OPQ = ∠OPT − ∠QPT = 90^∘ − 60^∘ = 30^∘
Since, OP = OQ as both are radius
∴∠OPQ = ∠OQP = 30^∘          (Angles opposite to equal sides are equal)
Now, In isosceles △POQ
∠POQ + ∠OPQ + ∠OQP = 180^∘       (Angle sum property of a triangle)
⇒ ∠POQ = 180^∘ − 30^∘ − 30^∘ = 120^∘
Now, ∠POQ + reflex ∠POQ = 360^∘         (Complete angle)
⇒ reflex ∠POQ = 360^∘ − 120^∘ = 240^∘
We know that the angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.
$∴ ∠ PRQ = \frac{1}{2} (reflex ∠ POQ) = 120 °$

Question 14:

In the given figure, PA and PB are two tangents to a circle with centre O, If ∠APB = 60^∘ then find the measure of ∠OAB

Answer:

Construction: Join OB

We know that the radius and tangent are perperpendular at their point of contact
∵∠OBP = ∠OAP = 90^∘
Now, In quadrilateral AOBP
∠AOB + ∠OBP + ∠APB + ∠OAP = 360^∘            [Angle sum property of a quadrilateral]
⇒ ∠AOB + 90^∘ + 60^∘ + 90^∘ = 360^∘
⇒ 240^∘ + ∠AOB = 360^∘
⇒ ∠AOB = 120⁰
Now, In isoceles triangle AOB
∠AOB + ∠OAB + ∠OBA = 180^∘            [Angle sum property of a triangle]
⇒ 120^∘ + 2∠OAB = 180^∘                          [∵∠OAB = ∠OBA ]
⇒ ∠OAB = 30^∘

Question 2:

A point P is 25 cm away from the centre of a circle and the length of tangent drawn from P to the circle is 24 cm. Find the radius of the circle.

Answer:

$Draw a circle and let P be a point such that OP = 25 cm . Let TP be the tangent, so that TP = 24 cm Join OT,where OT is radius . Now, tangent drawn from an external point is perpendicular to the radius at the point of contact . ∴ OT ⊥ PT In the right △ OTP, we have: {OP}^{2} = {OT}^{2} + {TP}^{2} [By Pythagoras’ theorem:] {OT}^{2} = \sqrt{{OP}^{2} - {TP}^{2}} = \sqrt{25^{2} - 24^{2}} = \sqrt{625 - 576} = \sqrt{49} =7 cm ∴ The length of the radius is 7 cm .$

Question 3:

Two concentric circles are of radii 6.5 cm and 2.5 cm. Find the length of the chord of the larger circle which touches the smaller circle

Answer:

We know that the radius and tangent are perperpendular at their point of contact
In right triangle AOP
AO² = OP² + PA²
⇒ (6.5)² = (2.5)² + PA²
⇒ PA² = 36
⇒ PA = 6 cm
Since, the perpendicular drawn from the centre bisect the chord.
∴ PA = PB = 6 cm
Now, AB = AP + PB = 6 + 6 = 12 cm
Hence, the length of the chord of the larger circle is 12 cm.

Question 4:

In the given figure, a circle inscribed in a triangle ABC, touches the sides AB, BC and AC at point D, E and F respectively. If AB = 12 cm, BC = 8 cm and AC = 10 cm, find the lengths of AD, BE and CF [CBSE 2013]

Answer:

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AD = AF, BD = BE and CE = CF
Now, AD + BD = 12 cm          …..(1)
AF + FC = 10 cm
⇒ AD + FC = 10 cm                    …..(2)
BE + EC = 8 cm
⇒ BD + FC = 8 cm                   …..(3)
Adding all these we get
AD + BD + AD + FC + BD + FC = 30
⇒2(AD + BD + FC) = 30
⇒AD + BD + FC = 15 cm           …..(4)
Solving (1) and (4), we get
FC = 3 cm
Solving (2) and (4), we get
BD = 5 cm
Solving (3) and (4), we get
and AD = 7 cm
∴ AD = AF = 7 cm, BD = BE = 5 cm and CE = CF = 3 cm

Question 5:

In the given figure, PA and PB are the tangents to a circle with centre O. Show that the points A, O, B, P are concyclic.

Answer:

$Here, O A = O B A nd O A ⊥ A P, O A ⊥ B P (S ince tangents drawn from an external point are perpendicular to the radius at the point of contact) ∴ ∠ O A P = 90^{0}, ∠ O B P = 90^{0} ∴ ∠ O A P + ∠ O B P = 90^{0} + 90^{0} = 180^{0} ∴ ∠ A O B + ∠ A P B = 180^{0} (S ince, ∠ O A P + ∠ O B P + ∠ A O B + ∠ A P B = 360^{0}) S um of opposite angle of a quadrilateral is 180 ° . Hence, A, O, B and P are concyclic .$

Question 6:

In the given figure, the chord AB of the larger of the two concentric circles, with centre O, touches the smaller circle at C. Prove that AC = CB.

Answer:

Construction: Join OA, OC and OB

We know that the radius and tangent are perperpendular at their point of contact
∴ ∠OCA = ∠OCB = 90^∘
Now, In △OCA and △OCB
∠OCA = ∠OCB = 90^∘
OA = OB (Radii of the larger circle)
OC = OC (Common)
By RHS congruency
△OCA ≅ △OCB
∴ CA = CB

Question 7:

From an external point P, tangents PA and PB are drawn to a circle with centre O. If CD is the tangent to the circle at a point E and PA = 14 cm, find the perimeter of ΔPCD.

Answer:

$Given, PA and PB are the tangents to a circle with centre O and CD is a tangent at E and PA = 14 cm . Tangents drawn from an external point are equal. ∴ PA = PB,CA = CE and DB = DE Perimeter of △ PCD = PC + CD + PD = (PA - CA) + (CE + DE) + (PB - DB) = (PA - C E) + (CE + DE) + (PB - DE) = (PA + PB) =2PA (∵ PA=PB) = (2 \times 14) cm =28 cm ∴ Perimeter of △ PCD = 28 cm .$

Question 8:

A circle is inscribed in ΔABC, touching AB, BC and AC at P, Q and R, respectively. If AB = 10 cm, AR = 7 cm and CR = 5 cm, find the length of BC.

Answer:

$Given, a circle inscribed in triangle ABC, such that the circle touches the sides of the triangle at P,Q,R . Tangents drawn to a circle from an external point are equal . ∴ AP = AR = 7 cm, CQ = CR = 5 cm . Now, BP = (AB - AP) = (10 - 7) = 3 cm ∴ BP = BQ = 3 cm ∴ BC = (BQ + QC) = > B C = 3 + 5 = > B C = 8 ∴ The length of BC is 8 cm .$

Question 9:

In the given figure, a circle touches all the four sides of a quadrilateral ABCD whose three sides are AB = 6 cm, BC = 7 cm and CD = 4 cm. Find AD.

Answer:

$Let the circle touch the sides of the quadrilateral AB,BC,CD and DA at P,Q,R and S respectively . Given, A B = 6 cm, B C = 7 cm and C D = 4 cm . Tangents drawn from an external point are equal. ∴ A P = A S, B P = B Q, C R = C Q and D R = D S \begin{array}{l} Now, A B + C D = (A P + B P) + (C R + D R) \\ = > A B + C D = (A S + B Q) + (C Q + D S) \\ = > A B + C D = (A S + D S) + (B Q + C Q) \\ = > A B + C D = A D + B C \\ = > A D = (A B + C D) - B C \\ = > A D = (6 + 4) - 7 \\ = > A D = 3 cm . \end{array} ∴ The length of A D is 3 cm .$

Question 10:

In the given figure, an isosceles triangle ABC with AB = AC, circumscribes a circle. Prove that the point of contact P bisects the base BC
[CBSE 2012]

Answer:

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AR = AQ, BR = BP and CP = CQ
Now, AB = AC
⇒ AR + RB = AQ + QC
⇒ AR + RB = AR + QC
⇒ RB = QC
⇒ BP = CP
Hence, P bisects BC at P.

Question 11:

In the given figure, O is the centre of two concentric circles of radii 4 cm and 6 cm respectively. PA and PB are tangents to the outer and inner circles, respectively. If PA = 10 cm, find the length of PB up to one decimal place.

Answer:

$Given, O is the centre of two concentric circles of radii O A = 6 cm and O B = 4 cm . P A and P B are the two tangents to the outer and inner circles respectively and P A = 10 cm . Now, tangent drawn from an external point is perpendicular to the radius at the point of contact . ∴ ∠ O A P = ∠ O B P = 90^{0} ∴ From right-angled △ O A P, O P^{2} = O A^{2} + P A^{2} \begin{array}{l} = > O P = \sqrt{O A^{2} + P A^{2}} \\ = > O P = \sqrt{6^{2} + 10^{2}} \\ = > O P = \sqrt{136} cm . \end{array} ∴ From right-angled △ O A P, O P^{2} = O B^{2} + P B^{2} \begin{array}{l} = > P B = \sqrt{O P^{2} - O B^{2}} \\ = > P B = \sqrt{136 - 16} \\ = > P B = \sqrt{120} cm \\ => P B = 10.9 cm . \end{array} ∴ The length of P B is 10.9 cm .$

Question 12:

In the given figure, a triangle ABC is drawn to circumscribe a circle of radius 3 cm such that the segments BD and DC into which BC is divided by the point of contact D are, of lengths 6 cm and 9 cm respectively. If the area of △ABC = 54 cm² then find the lengths of sides of AB and AC. [CBSE 2011, ’15]

Answer:

Construction: Join OA, OB, OC, OE ⊥ AB at E and OF ⊥ AC at F

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AE = AF, BD = BE = 6 cm and CD = CF = 9 cm
Now,
$Area (△ ABC) = Area (△ BOC) + Area (△ AOB) + Area (△ AOC) \Rightarrow 54 = \frac{1}{2} \times BC \times OD + \frac{1}{2} \times AB \times OE + \frac{1}{2} \times AC \times OF \Rightarrow 108 = 15 \times 3 + (6 + x) \times 3 + (9 + x) \times 3 \Rightarrow 36 = 15 + 6 + x + 9 + x$

$\Rightarrow 36 = 30 + 2 x \Rightarrow 2 x = 6 \Rightarrow x = 3 cm$
∴ AB = 6 + 3 = 9cm and AC = 9 + 3 = 12 cm

Question 13:

PQ is a chord of length 4.8 cm of a circle of radius 3 cm. The tangent at P and Q intersect at a point T as shown in the figure. Find the length of TP [CBSE 2013C]

Answer:

Let TR = y and TP = x
We know that the perpendicular drawn from the centre to the chord bisects it.
∴ PR = RQ
Now, PR + RQ = 4.8
⇒ PR + PR = 4.8
⇒ PR = 2.4
Now, in right triangle POR
By Using Pyhthagoras theorem, we have
PO² = OR² + PR²
⇒ 3² = OR² + (2.4)²
⇒ OR² = 3.24
⇒ OR = 1.8
Now, in right triangle TPR
By Using Pyhthagoras theorem, we have
TP² = TR² + PR²
⇒ x² = y² + (2.4)²
⇒ x² = y² + 5.76 …..(1)
Again, in right triangle TPQ
By Using Pyhthagoras theorem, we have
TO² = TP² + PO²
⇒ (y + 1.8)² = x² + 3²
⇒ y² + 3.6y + 3.24 = x² + 9
⇒ y² + 3.6y = x² + 5.76 …..(2)
Solving (1) and (2), we get
x = 4 cm and y = 3.2 cm
∴ TP = 4 cm

Question 14:

Prove that the line joining the points of contact of two parallel tangents of a circle passes through its centre. [CBSE 2014]

Answer:

Suppose CD and AB are two parallel tangents of a circle with centre O
Construction: Draw a line parallel to CD passing through O i.e, OP
We know that the radius and tangent are perperpendular at their point of contact.
∠OQC = ∠ORA = 90^∘
Now, ∠OQC + ∠POQ = 180^∘ (co-interior angles)
⇒ ∠POQ = 180^∘ − 90^∘ = 90^∘
Similarly, Now, ∠ORA + ∠POR = 180^∘ (co-interior angles)
⇒ ∠POR = 180^∘ − 90^∘ = 90^∘
Now, ∠POR + ∠POQ = 90^∘ + 90^∘ = 180^∘
Since, ∠POR and ∠POQ are linear pair angles whose sum is 180^∘
Hence, QR is a straight line passing through centre O.

Question 15:

In the given figure, a circle with centre O, is inscribed in a quadrilateral ABCD such that it touches the side BC, AB, AD and CD at points P, Q, R and S respectively. If AB = 29 cm, AD = 23 cm, ∠B = 90^∘ and DS = 5 cm then find the radius of the circle. [CBSE 2008, 13]

Answer:

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
DS = DR, AR = AQ
Now, AD = 23 cm
⇒ AR + RD = 23
⇒ AR = 23 − RD
⇒ AR = 23 − 5 [∵ DS = DR = 5]
⇒ AR = 18 cm
Again, AB = 29 cm
⇒ AQ + QB = 29
⇒ QB = 29 − AQ
⇒ QB = 29 − 18 [∵ AR = AQ = 18]
⇒ QB = 11 cm
Since all the angles are in a quadrilateral BQOP are right angles and OP = BQ.
Hence, BQOP is a square.
We know that all the sides of square are equal.
Therefore, BQ = PO = 11 cm
Hence, the radius of the circle is 11 cm.

Question 16:

In the given figure, O is the centre of the circle and TP is the tangent to the circle from an external point T. If ∠PBT = 30^∘ , prove that
BA : AT = 2 : 1 [CBSE 2015]

Answer:

AB is the chord passing through the centre
So, AB is the diameter
Since, angle in a semi circle is a right angle
∴∠APB = 90^∘
By using alternate segment theorem
We have ∠APB = ∠PAT = 30^∘
Now, in △APB
∠BAP + ∠APB + ∠BAP = 180^∘ (Angle sum property of triangle)
⇒ ∠BAP = 180^∘ − 90^∘ − 30^∘ = 60^∘
Now, ∠BAP = ∠APT + ∠PTA (Exterior angle property)
⇒ 60^∘ = 30^∘ + ∠PTA
⇒ ∠PTA = 60^∘ − 30^∘ = 30^∘
We know that sides opposite to equal angles are equal.
∴ AP = AT
In right triangle ABP
$\sin ∠ ABP = \frac{AP}{BA} \Rightarrow \sin 30 ° = \frac{AT}{BA} \Rightarrow \frac{1}{2} = \frac{AT}{BA}$
∴ BA : AT = 2 : 1

Question 1:

In the adjoining figure, a circle touches all the four sides of a quadrilateral ABCD whose sides are AB = 6 cm, BC = 9 cm and CD = 8 cm. Find the length of AD [CBSE 2011]

Answer:

We know that when a quadrilateral circumscribes a circle then sum of opposites sides is equal to the sum of other opposite sides.
∴ AB + CD = AD + BC
⇒6 + 8 = AD + 9
⇒ AD = 5 cm

Question 2:

In the given figure, PA and PB are two tangents to the circle with centre O. If ∠APB = 50^∘ then what is the measure of ∠OAB is [CBSE 2015]

Answer:

Construction: Join OB

We know that the radius and tangent are perperpendular at their point of contact
∵∠OBP = ∠OAP = 90^∘
Now, In quadrilateral AOBP
∠AOB + ∠OBP + ∠APB + ∠OAP = 360^∘            [Angle sum property of a quadrilateral]
⇒ ∠AOB + 90^∘ + 50^∘+ 90^∘ = 360^∘
⇒ 230^∘+ ∠BOC = 360^∘
⇒ ∠AOB = 130^∘
Now, In isoceles triangle AOB
∠AOB + ∠OAB + ∠OBA = 180^∘            [Angle sum property of a triangle]
⇒ 130^∘ + 2∠OAB = 180⁰                          [∵∠OAB = ∠OBA ]
⇒ ∠OAB = 25^∘

Question 3:

In the given figure, O is the centre of a circle PT and PQ are tangents to the circle from an external point P. If ∠TPQ = 70^∘ then ∠TRQ
[CBSE 2015]

Answer:

Construction: Join OQ and OT

We know that the radius and tangent are perperpendular at their point of contact
∵∠OTP = ∠OQP = 90^∘
Now, In quadrilateral OQPT
∠QOT + ∠OTP + ∠OQP + ∠TPQ = 360^∘ [Angle sum property of a quadrilateral]
⇒ ∠QOT + 90^∘ + 90^∘ + 70^∘ = 360^∘
⇒ 250^∘ + ∠QOT = 360^∘
⇒ ∠QOT = 110^∘
We know that the angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.
$∴ ∠ TRQ = \frac{1}{2} (∠ QOT) = 55 °$

Question 4:

In the given figure, common tangents AB and CD to the two circle with centres O₁ and O₂ intersect at E. Prove that AB = CD [CBSE 2014]

Answer:

We know that tangent segments to a circle from the same external point are congruent.
So, we have
EA = EC for the circle having centre O₁
and
ED = EB for the circle having centre O₁
Now, Adding ED on both sides in EA = EC, we get
EA + ED = EC + ED
⇒EA + EB = EC + ED
⇒AB = CD

Question 5:

If PT is a tangent to a circle with centre O and PQ is a chord of the circle such that ∠QPT = 70^∘ then find the measure of ∠POQ

Answer:

We know that the radius and tangent are perperpendular at their point of contact.
∴∠OPT = 90^∘
Now, ∠OPQ = ∠OPT − ∠TPQ = 90^∘ − 70^∘ = 20^∘
Since, OP = OQ as both are radius
∴∠OPQ = ∠OQP = 20^∘ (Angles opposite to equal sides are equal)
Now, In isosceles △POQ
∠POQ + ∠OPQ + ∠OQP = 180^∘ (Angle sum property of a triangle)
⇒ ∠POQ = 180^∘ − 20^∘ − 20^∘ = 140^∘

Question 6:

In the given figure, a triangle ABC is drawn to circumscribe a circle of radius 2 cm such that the segments BD and DC into which BC is divided by the point of contact D are, of lengths 4 cm and 3 cm respectively. If the area of △ABC = 21 cm² then find the lengths of sides of AB and AC. [CBSE 2011]

Answer:

Construction: Join OA, OB, OC, OE ⊥ AB at E and OF ⊥ AC at F

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AE = AF, BD = BE = 4 cm and CD = CF = 3 cm
Now,
$Area (△ ABC) = Area (△ BOC) + Area (△ AOB) + Area (△ AOC) \Rightarrow 21 = \frac{1}{2} \times BC \times OD + \frac{1}{2} \times AB \times OE + \frac{1}{2} \times AC \times OF \Rightarrow 42 = 7 \times 2 + (4 + x) \times 2 + (3 + x) \times 2 \Rightarrow 21 = 7 + 4 + x + 3 + x$

$\Rightarrow 21 = 14 + 2 x \Rightarrow 2 x = 7 \Rightarrow x = 3.5 cm$
∴ AB = 4 + 3.5 = 7.5 cm and AC = 3 + 3.5 = 6.5 cm

Question 7:

Two concentric circles are of radii 5 cm and 3 cm, respectively. Find the length of the chord of the larger circle that touches the smaller circle.

Answer:

Given: Two circles have the same centre O and AB is a chord of the larger circle touching the
smaller circle at C; also, OA=5 cm and OC=3 cm.
$In ∆ O A C, O A^{2} = O C^{2} + A C^{2} ∴ A C^{2} = O A^{2} - O C^{2} \Rightarrow A C^{2} = 5^{2} - 3^{2} \Rightarrow A C^{2} = 25 - 9 \Rightarrow A C^{2} = 16 \Rightarrow A C = 4 cm ∴ A B = 2 A C (since perpendicular drawn from the centre of the circle bisects the chord) ∴ A B = 2 \times 4 = 8 cm$
The length of the chord of the larger circle is 8 cm.

Question 8:

Prove that the perpendicular at the point of contact of the tangent to a circle passes through the centre.

Answer:

Let AB be the tangent to the circle at point P with centre O.
To prove: PQ passes through the point O.
Construction: Join OP.
Through O, draw a straight line CD parallel to the tangent AB.
Proof: Suppose that PQ doesn’t passes through point O.
PQ intersect CD at R and also intersect AB at P.
AS, CD ∥ AB, PQ is the line of intersection,
∠ORP = ∠RPA (Alternate interior angles)
but also,
∠RPA = 90^∘ (OP ⊥ AB)
⇒ ∠ORP = 90^∘
∠ROP + ∠OPA = 180^∘ (Co interior angles)
⇒∠ROP + 90^∘ = 180^∘
⇒∠ROP = 90^∘
Thus, the ΔORP has 2 right angles i.e. ∠ORP and ∠ROP which is not possible.
Hence, our supposition is wrong.
∴ PQ passes through the point O.

Question 9:

In the given figure, two tangents RQ and RP are drawn from an external point R to the circle with centre O. If ∠PRQ = 120^∘ then prove that OR = PR + RQ

Answer:

Construction: Join PO and OQ
In △POR and △QOR
OP = OQ   (Radii)
RP = RQ    (Tangents from the external point are congruent)
OR = OR    (Common)
By SSS congruency, △POR ≅ △QOR
∠PRO = ∠QRO    (C.P.C.T)
Now, ∠PRO + ∠QRO = ∠PRQ
⇒ 2∠PRO = 120^∘
⇒ ∠PRO = 60^∘
Now, In △POR
$\cos 60^{°} = \frac{PR}{OR} \Rightarrow \frac{1}{2} = \frac{PR}{OR} \Rightarrow OR = 2 PR \Rightarrow OR = PR + PR \Rightarrow OR = PR + RQ$

Question 10:

In the given figure, a circle inscribed in a triangle ABC, touches the sides AB, BC and AC at point D, E and F respectively. If AB = 14 cm, BC = 8 cm and AC = 12 cm. Find the lengths of AD, BE and CF [CBSE 2013]

Answer:

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AD = AF, BD = BE and CE = CF
Now, AD + BD = 14 cm          …..(1)
AF + FC = 12 cm
⇒ AD + FC = 12 cm                    …..(2)
BE + EC = 8 cm
⇒ BD + FC = 8 cm                   …..(3)
Adding all these we get
AD + BD + AD + FC + BD + FC = 34
⇒2(AD + BD + FC) = 34
⇒AD + BD + FC = 17 cm           …..(4)
Solving (1) and (4), we get
FC = 3 cm
Solving (2) and (4), we get
BD = 5 cm = BE
Solving (3) and (4), we get
and AD = 9 cm

Question 11:

In the given figure, O is the centre of the circle. PA and PB are tangents. Show that AOBP is a cyclic quadrilateral [CBSE 2014]

Answer:

We know that the radius and tangent are perpendicular at their point of contact
∵∠OBP = ∠OAP = 90^∘
Now, In quadrilateral AOBP
∠APB + ∠AOB + ∠OBP + ∠OAP = 360^∘ [Angle sum property of a quadrilateral]
⇒ ∠APB + ∠AOB + 90^∘ + 90^∘= 360^∘
⇒ ∠APB + ∠AOB = 180^∘
Also, ∠OBP + ∠OAP = 180^∘
Since, the sum of the opposite angles of the quadrilateral is 180^∘
Hence, AOBP is a cyclic quadrilateral.

Question 12:

In two concentric circles, a chord of length 8 cm of the larger circle touches the smaller circle. If the radius of the larger circle is 5 cm then Find the radius of the smaller circle. [CBSE 2013C]

Answer:

We know that the radius and tangent are perperpendular at their point of contact
Since, the perpendicular drawn from the centre bisect the chord.
∴ AP = PB = $\frac{AB}{2}$ = 4 cm
In right triangle AOP
AO² = OP² + PA²
⇒ 5² = OP² + 4²
⇒ OP² = 9
⇒ OP = 3 cm
Hence, the radius of the smaller circle is 3 cm.

Question 13:

In the given figure, PQ is a chord of a circle with centre O and PT is a tangent. If ∠QPT = 60^∘ , find ∠PRQ

Answer:

We know that the radius and tangent are perperpendular at their point of contact.
∴∠OPT = 90^∘
Now, ∠OPQ = ∠OPT − ∠QPT = 90^∘ − 60^∘ = 30^∘
Since, OP = OQ as both are radius
∴∠OPQ = ∠OQP = 30^∘          (Angles opposite to equal sides are equal)
Now, In isosceles △POQ
∠POQ + ∠OPQ + ∠OQP = 180^∘       (Angle sum property of a triangle)
⇒ ∠POQ = 180^∘ − 30^∘ − 30^∘ = 120^∘
Now, ∠POQ + reflex ∠POQ = 360^∘         (Complete angle)
⇒ reflex ∠POQ = 360^∘ − 120^∘ = 240^∘
We know that the angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.
$∴ ∠ PRQ = \frac{1}{2} (reflex ∠ POQ) = 120 °$

Question 14:

In the given figure, PA and PB are two tangents to a circle with centre O, If ∠APB = 60^∘ then find the measure of ∠OAB

Answer:

Construction: Join OB

We know that the radius and tangent are perperpendular at their point of contact
∵∠OBP = ∠OAP = 90^∘
Now, In quadrilateral AOBP
∠AOB + ∠OBP + ∠APB + ∠OAP = 360^∘            [Angle sum property of a quadrilateral]
⇒ ∠AOB + 90^∘ + 60^∘ + 90^∘ = 360^∘
⇒ 240^∘ + ∠AOB = 360^∘
⇒ ∠AOB = 120⁰
Now, In isoceles triangle AOB
∠AOB + ∠OAB + ∠OBA = 180^∘            [Angle sum property of a triangle]
⇒ 120^∘ + 2∠OAB = 180^∘                          [∵∠OAB = ∠OBA ]
⇒ ∠OAB = 30^∘

Question 2:

A point P is 25 cm away from the centre of a circle and the length of tangent drawn from P to the circle is 24 cm. Find the radius of the circle.

Answer:

$Draw a circle and let P be a point such that OP = 25 cm . Let TP be the tangent, so that TP = 24 cm Join OT,where OT is radius . Now, tangent drawn from an external point is perpendicular to the radius at the point of contact . ∴ OT ⊥ PT In the right △ OTP, we have: {OP}^{2} = {OT}^{2} + {TP}^{2} [By Pythagoras’ theorem:] {OT}^{2} = \sqrt{{OP}^{2} - {TP}^{2}} = \sqrt{25^{2} - 24^{2}} = \sqrt{625 - 576} = \sqrt{49} =7 cm ∴ The length of the radius is 7 cm .$

Question 3:

Two concentric circles are of radii 6.5 cm and 2.5 cm. Find the length of the chord of the larger circle which touches the smaller circle

Answer:

We know that the radius and tangent are perperpendular at their point of contact
In right triangle AOP
AO² = OP² + PA²
⇒ (6.5)² = (2.5)² + PA²
⇒ PA² = 36
⇒ PA = 6 cm
Since, the perpendicular drawn from the centre bisect the chord.
∴ PA = PB = 6 cm
Now, AB = AP + PB = 6 + 6 = 12 cm
Hence, the length of the chord of the larger circle is 12 cm.

Question 4:

In the given figure, a circle inscribed in a triangle ABC, touches the sides AB, BC and AC at point D, E and F respectively. If AB = 12 cm, BC = 8 cm and AC = 10 cm, find the lengths of AD, BE and CF [CBSE 2013]

Answer:

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AD = AF, BD = BE and CE = CF
Now, AD + BD = 12 cm          …..(1)
AF + FC = 10 cm
⇒ AD + FC = 10 cm                    …..(2)
BE + EC = 8 cm
⇒ BD + FC = 8 cm                   …..(3)
Adding all these we get
AD + BD + AD + FC + BD + FC = 30
⇒2(AD + BD + FC) = 30
⇒AD + BD + FC = 15 cm           …..(4)
Solving (1) and (4), we get
FC = 3 cm
Solving (2) and (4), we get
BD = 5 cm
Solving (3) and (4), we get
and AD = 7 cm
∴ AD = AF = 7 cm, BD = BE = 5 cm and CE = CF = 3 cm

Question 5:

In the given figure, PA and PB are the tangents to a circle with centre O. Show that the points A, O, B, P are concyclic.

Answer:

$Here, O A = O B A nd O A ⊥ A P, O A ⊥ B P (S ince tangents drawn from an external point are perpendicular to the radius at the point of contact) ∴ ∠ O A P = 90^{0}, ∠ O B P = 90^{0} ∴ ∠ O A P + ∠ O B P = 90^{0} + 90^{0} = 180^{0} ∴ ∠ A O B + ∠ A P B = 180^{0} (S ince, ∠ O A P + ∠ O B P + ∠ A O B + ∠ A P B = 360^{0}) S um of opposite angle of a quadrilateral is 180 ° . Hence, A, O, B and P are concyclic .$

Question 6:

In the given figure, the chord AB of the larger of the two concentric circles, with centre O, touches the smaller circle at C. Prove that AC = CB.

Answer:

Construction: Join OA, OC and OB

We know that the radius and tangent are perperpendular at their point of contact
∴ ∠OCA = ∠OCB = 90^∘
Now, In △OCA and △OCB
∠OCA = ∠OCB = 90^∘
OA = OB (Radii of the larger circle)
OC = OC (Common)
By RHS congruency
△OCA ≅ △OCB
∴ CA = CB

Question 7:

From an external point P, tangents PA and PB are drawn to a circle with centre O. If CD is the tangent to the circle at a point E and PA = 14 cm, find the perimeter of ΔPCD.

Answer:

$Given, PA and PB are the tangents to a circle with centre O and CD is a tangent at E and PA = 14 cm . Tangents drawn from an external point are equal. ∴ PA = PB,CA = CE and DB = DE Perimeter of △ PCD = PC + CD + PD = (PA - CA) + (CE + DE) + (PB - DB) = (PA - C E) + (CE + DE) + (PB - DE) = (PA + PB) =2PA (∵ PA=PB) = (2 \times 14) cm =28 cm ∴ Perimeter of △ PCD = 28 cm .$

Question 8:

A circle is inscribed in ΔABC, touching AB, BC and AC at P, Q and R, respectively. If AB = 10 cm, AR = 7 cm and CR = 5 cm, find the length of BC.

Answer:

$Given, a circle inscribed in triangle ABC, such that the circle touches the sides of the triangle at P,Q,R . Tangents drawn to a circle from an external point are equal . ∴ AP = AR = 7 cm, CQ = CR = 5 cm . Now, BP = (AB - AP) = (10 - 7) = 3 cm ∴ BP = BQ = 3 cm ∴ BC = (BQ + QC) = > B C = 3 + 5 = > B C = 8 ∴ The length of BC is 8 cm .$

Question 9:

In the given figure, a circle touches all the four sides of a quadrilateral ABCD whose three sides are AB = 6 cm, BC = 7 cm and CD = 4 cm. Find AD.

Answer:

$Let the circle touch the sides of the quadrilateral AB,BC,CD and DA at P,Q,R and S respectively . Given, A B = 6 cm, B C = 7 cm and C D = 4 cm . Tangents drawn from an external point are equal. ∴ A P = A S, B P = B Q, C R = C Q and D R = D S \begin{array}{l} Now, A B + C D = (A P + B P) + (C R + D R) \\ = > A B + C D = (A S + B Q) + (C Q + D S) \\ = > A B + C D = (A S + D S) + (B Q + C Q) \\ = > A B + C D = A D + B C \\ = > A D = (A B + C D) - B C \\ = > A D = (6 + 4) - 7 \\ = > A D = 3 cm . \end{array} ∴ The length of A D is 3 cm .$

Question 10:

In the given figure, an isosceles triangle ABC with AB = AC, circumscribes a circle. Prove that the point of contact P bisects the base BC
[CBSE 2012]

Answer:

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AR = AQ, BR = BP and CP = CQ
Now, AB = AC
⇒ AR + RB = AQ + QC
⇒ AR + RB = AR + QC
⇒ RB = QC
⇒ BP = CP
Hence, P bisects BC at P.

Question 11:

In the given figure, O is the centre of two concentric circles of radii 4 cm and 6 cm respectively. PA and PB are tangents to the outer and inner circles, respectively. If PA = 10 cm, find the length of PB up to one decimal place.

Answer:

$Given, O is the centre of two concentric circles of radii O A = 6 cm and O B = 4 cm . P A and P B are the two tangents to the outer and inner circles respectively and P A = 10 cm . Now, tangent drawn from an external point is perpendicular to the radius at the point of contact . ∴ ∠ O A P = ∠ O B P = 90^{0} ∴ From right-angled △ O A P, O P^{2} = O A^{2} + P A^{2} \begin{array}{l} = > O P = \sqrt{O A^{2} + P A^{2}} \\ = > O P = \sqrt{6^{2} + 10^{2}} \\ = > O P = \sqrt{136} cm . \end{array} ∴ From right-angled △ O A P, O P^{2} = O B^{2} + P B^{2} \begin{array}{l} = > P B = \sqrt{O P^{2} - O B^{2}} \\ = > P B = \sqrt{136 - 16} \\ = > P B = \sqrt{120} cm \\ => P B = 10.9 cm . \end{array} ∴ The length of P B is 10.9 cm .$

Question 12:

In the given figure, a triangle ABC is drawn to circumscribe a circle of radius 3 cm such that the segments BD and DC into which BC is divided by the point of contact D are, of lengths 6 cm and 9 cm respectively. If the area of △ABC = 54 cm² then find the lengths of sides of AB and AC. [CBSE 2011, ’15]

Answer:

Construction: Join OA, OB, OC, OE ⊥ AB at E and OF ⊥ AC at F

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AE = AF, BD = BE = 6 cm and CD = CF = 9 cm
Now,
$Area (△ ABC) = Area (△ BOC) + Area (△ AOB) + Area (△ AOC) \Rightarrow 54 = \frac{1}{2} \times BC \times OD + \frac{1}{2} \times AB \times OE + \frac{1}{2} \times AC \times OF \Rightarrow 108 = 15 \times 3 + (6 + x) \times 3 + (9 + x) \times 3 \Rightarrow 36 = 15 + 6 + x + 9 + x$

$\Rightarrow 36 = 30 + 2 x \Rightarrow 2 x = 6 \Rightarrow x = 3 cm$
∴ AB = 6 + 3 = 9cm and AC = 9 + 3 = 12 cm

Question 13:

PQ is a chord of length 4.8 cm of a circle of radius 3 cm. The tangent at P and Q intersect at a point T as shown in the figure. Find the length of TP [CBSE 2013C]

Answer:

Let TR = y and TP = x
We know that the perpendicular drawn from the centre to the chord bisects it.
∴ PR = RQ
Now, PR + RQ = 4.8
⇒ PR + PR = 4.8
⇒ PR = 2.4
Now, in right triangle POR
By Using Pyhthagoras theorem, we have
PO² = OR² + PR²
⇒ 3² = OR² + (2.4)²
⇒ OR² = 3.24
⇒ OR = 1.8
Now, in right triangle TPR
By Using Pyhthagoras theorem, we have
TP² = TR² + PR²
⇒ x² = y² + (2.4)²
⇒ x² = y² + 5.76 …..(1)
Again, in right triangle TPQ
By Using Pyhthagoras theorem, we have
TO² = TP² + PO²
⇒ (y + 1.8)² = x² + 3²
⇒ y² + 3.6y + 3.24 = x² + 9
⇒ y² + 3.6y = x² + 5.76 …..(2)
Solving (1) and (2), we get
x = 4 cm and y = 3.2 cm
∴ TP = 4 cm

Question 14:

Prove that the line joining the points of contact of two parallel tangents of a circle passes through its centre. [CBSE 2014]

Answer:

Suppose CD and AB are two parallel tangents of a circle with centre O
Construction: Draw a line parallel to CD passing through O i.e, OP
We know that the radius and tangent are perperpendular at their point of contact.
∠OQC = ∠ORA = 90^∘
Now, ∠OQC + ∠POQ = 180^∘ (co-interior angles)
⇒ ∠POQ = 180^∘ − 90^∘ = 90^∘
Similarly, Now, ∠ORA + ∠POR = 180^∘ (co-interior angles)
⇒ ∠POR = 180^∘ − 90^∘ = 90^∘
Now, ∠POR + ∠POQ = 90^∘ + 90^∘ = 180^∘
Since, ∠POR and ∠POQ are linear pair angles whose sum is 180^∘
Hence, QR is a straight line passing through centre O.

Question 15:

In the given figure, a circle with centre O, is inscribed in a quadrilateral ABCD such that it touches the side BC, AB, AD and CD at points P, Q, R and S respectively. If AB = 29 cm, AD = 23 cm, ∠B = 90^∘ and DS = 5 cm then find the radius of the circle. [CBSE 2008, 13]

Answer:

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
DS = DR, AR = AQ
Now, AD = 23 cm
⇒ AR + RD = 23
⇒ AR = 23 − RD
⇒ AR = 23 − 5 [∵ DS = DR = 5]
⇒ AR = 18 cm
Again, AB = 29 cm
⇒ AQ + QB = 29
⇒ QB = 29 − AQ
⇒ QB = 29 − 18 [∵ AR = AQ = 18]
⇒ QB = 11 cm
Since all the angles are in a quadrilateral BQOP are right angles and OP = BQ.
Hence, BQOP is a square.
We know that all the sides of square are equal.
Therefore, BQ = PO = 11 cm
Hence, the radius of the circle is 11 cm.

Question 16:

In the given figure, O is the centre of the circle and TP is the tangent to the circle from an external point T. If ∠PBT = 30^∘ , prove that
BA : AT = 2 : 1 [CBSE 2015]

Answer:

AB is the chord passing through the centre
So, AB is the diameter
Since, angle in a semi circle is a right angle
∴∠APB = 90^∘
By using alternate segment theorem
We have ∠APB = ∠PAT = 30^∘
Now, in △APB
∠BAP + ∠APB + ∠BAP = 180^∘ (Angle sum property of triangle)
⇒ ∠BAP = 180^∘ − 90^∘ − 30^∘ = 60^∘
Now, ∠BAP = ∠APT + ∠PTA (Exterior angle property)
⇒ 60^∘ = 30^∘ + ∠PTA
⇒ ∠PTA = 60^∘ − 30^∘ = 30^∘
We know that sides opposite to equal angles are equal.
∴ AP = AT
In right triangle ABP
$\sin ∠ ABP = \frac{AP}{BA} \Rightarrow \sin 30 ° = \frac{AT}{BA} \Rightarrow \frac{1}{2} = \frac{AT}{BA}$
∴ BA : AT = 2 : 1

Question 1:

In the adjoining figure, a circle touches all the four sides of a quadrilateral ABCD whose sides are AB = 6 cm, BC = 9 cm and CD = 8 cm. Find the length of AD [CBSE 2011]

Answer:

We know that when a quadrilateral circumscribes a circle then sum of opposites sides is equal to the sum of other opposite sides.
∴ AB + CD = AD + BC
⇒6 + 8 = AD + 9
⇒ AD = 5 cm

Question 2:

In the given figure, PA and PB are two tangents to the circle with centre O. If ∠APB = 50^∘ then what is the measure of ∠OAB is [CBSE 2015]

Answer:

Construction: Join OB

We know that the radius and tangent are perperpendular at their point of contact
∵∠OBP = ∠OAP = 90^∘
Now, In quadrilateral AOBP
∠AOB + ∠OBP + ∠APB + ∠OAP = 360^∘            [Angle sum property of a quadrilateral]
⇒ ∠AOB + 90^∘ + 50^∘+ 90^∘ = 360^∘
⇒ 230^∘+ ∠BOC = 360^∘
⇒ ∠AOB = 130^∘
Now, In isoceles triangle AOB
∠AOB + ∠OAB + ∠OBA = 180^∘            [Angle sum property of a triangle]
⇒ 130^∘ + 2∠OAB = 180⁰                          [∵∠OAB = ∠OBA ]
⇒ ∠OAB = 25^∘

Question 3:

In the given figure, O is the centre of a circle PT and PQ are tangents to the circle from an external point P. If ∠TPQ = 70^∘ then ∠TRQ
[CBSE 2015]

Answer:

Construction: Join OQ and OT

We know that the radius and tangent are perperpendular at their point of contact
∵∠OTP = ∠OQP = 90^∘
Now, In quadrilateral OQPT
∠QOT + ∠OTP + ∠OQP + ∠TPQ = 360^∘ [Angle sum property of a quadrilateral]
⇒ ∠QOT + 90^∘ + 90^∘ + 70^∘ = 360^∘
⇒ 250^∘ + ∠QOT = 360^∘
⇒ ∠QOT = 110^∘
We know that the angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.
$∴ ∠ TRQ = \frac{1}{2} (∠ QOT) = 55 °$

Question 4:

In the given figure, common tangents AB and CD to the two circle with centres O₁ and O₂ intersect at E. Prove that AB = CD [CBSE 2014]

Answer:

We know that tangent segments to a circle from the same external point are congruent.
So, we have
EA = EC for the circle having centre O₁
and
ED = EB for the circle having centre O₁
Now, Adding ED on both sides in EA = EC, we get
EA + ED = EC + ED
⇒EA + EB = EC + ED
⇒AB = CD

Question 5:

If PT is a tangent to a circle with centre O and PQ is a chord of the circle such that ∠QPT = 70^∘ then find the measure of ∠POQ

Answer:

We know that the radius and tangent are perperpendular at their point of contact.
∴∠OPT = 90^∘
Now, ∠OPQ = ∠OPT − ∠TPQ = 90^∘ − 70^∘ = 20^∘
Since, OP = OQ as both are radius
∴∠OPQ = ∠OQP = 20^∘ (Angles opposite to equal sides are equal)
Now, In isosceles △POQ
∠POQ + ∠OPQ + ∠OQP = 180^∘ (Angle sum property of a triangle)
⇒ ∠POQ = 180^∘ − 20^∘ − 20^∘ = 140^∘

Question 6:

In the given figure, a triangle ABC is drawn to circumscribe a circle of radius 2 cm such that the segments BD and DC into which BC is divided by the point of contact D are, of lengths 4 cm and 3 cm respectively. If the area of △ABC = 21 cm² then find the lengths of sides of AB and AC. [CBSE 2011]

Answer:

Construction: Join OA, OB, OC, OE ⊥ AB at E and OF ⊥ AC at F

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AE = AF, BD = BE = 4 cm and CD = CF = 3 cm
Now,
$Area (△ ABC) = Area (△ BOC) + Area (△ AOB) + Area (△ AOC) \Rightarrow 21 = \frac{1}{2} \times BC \times OD + \frac{1}{2} \times AB \times OE + \frac{1}{2} \times AC \times OF \Rightarrow 42 = 7 \times 2 + (4 + x) \times 2 + (3 + x) \times 2 \Rightarrow 21 = 7 + 4 + x + 3 + x$

$\Rightarrow 21 = 14 + 2 x \Rightarrow 2 x = 7 \Rightarrow x = 3.5 cm$
∴ AB = 4 + 3.5 = 7.5 cm and AC = 3 + 3.5 = 6.5 cm

Question 7:

Two concentric circles are of radii 5 cm and 3 cm, respectively. Find the length of the chord of the larger circle that touches the smaller circle.

Answer:

Given: Two circles have the same centre O and AB is a chord of the larger circle touching the
smaller circle at C; also, OA=5 cm and OC=3 cm.
$In ∆ O A C, O A^{2} = O C^{2} + A C^{2} ∴ A C^{2} = O A^{2} - O C^{2} \Rightarrow A C^{2} = 5^{2} - 3^{2} \Rightarrow A C^{2} = 25 - 9 \Rightarrow A C^{2} = 16 \Rightarrow A C = 4 cm ∴ A B = 2 A C (since perpendicular drawn from the centre of the circle bisects the chord) ∴ A B = 2 \times 4 = 8 cm$
The length of the chord of the larger circle is 8 cm.

Question 8:

Prove that the perpendicular at the point of contact of the tangent to a circle passes through the centre.

Answer:

Let AB be the tangent to the circle at point P with centre O.
To prove: PQ passes through the point O.
Construction: Join OP.
Through O, draw a straight line CD parallel to the tangent AB.
Proof: Suppose that PQ doesn’t passes through point O.
PQ intersect CD at R and also intersect AB at P.
AS, CD ∥ AB, PQ is the line of intersection,
∠ORP = ∠RPA (Alternate interior angles)
but also,
∠RPA = 90^∘ (OP ⊥ AB)
⇒ ∠ORP = 90^∘
∠ROP + ∠OPA = 180^∘ (Co interior angles)
⇒∠ROP + 90^∘ = 180^∘
⇒∠ROP = 90^∘
Thus, the ΔORP has 2 right angles i.e. ∠ORP and ∠ROP which is not possible.
Hence, our supposition is wrong.
∴ PQ passes through the point O.

Question 9:

In the given figure, two tangents RQ and RP are drawn from an external point R to the circle with centre O. If ∠PRQ = 120^∘ then prove that OR = PR + RQ

Answer:

Construction: Join PO and OQ
In △POR and △QOR
OP = OQ   (Radii)
RP = RQ    (Tangents from the external point are congruent)
OR = OR    (Common)
By SSS congruency, △POR ≅ △QOR
∠PRO = ∠QRO    (C.P.C.T)
Now, ∠PRO + ∠QRO = ∠PRQ
⇒ 2∠PRO = 120^∘
⇒ ∠PRO = 60^∘
Now, In △POR
$\cos 60^{°} = \frac{PR}{OR} \Rightarrow \frac{1}{2} = \frac{PR}{OR} \Rightarrow OR = 2 PR \Rightarrow OR = PR + PR \Rightarrow OR = PR + RQ$

Question 10:

In the given figure, a circle inscribed in a triangle ABC, touches the sides AB, BC and AC at point D, E and F respectively. If AB = 14 cm, BC = 8 cm and AC = 12 cm. Find the lengths of AD, BE and CF [CBSE 2013]

Answer:

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AD = AF, BD = BE and CE = CF
Now, AD + BD = 14 cm          …..(1)
AF + FC = 12 cm
⇒ AD + FC = 12 cm                    …..(2)
BE + EC = 8 cm
⇒ BD + FC = 8 cm                   …..(3)
Adding all these we get
AD + BD + AD + FC + BD + FC = 34
⇒2(AD + BD + FC) = 34
⇒AD + BD + FC = 17 cm           …..(4)
Solving (1) and (4), we get
FC = 3 cm
Solving (2) and (4), we get
BD = 5 cm = BE
Solving (3) and (4), we get
and AD = 9 cm

Question 11:

In the given figure, O is the centre of the circle. PA and PB are tangents. Show that AOBP is a cyclic quadrilateral [CBSE 2014]

Answer:

We know that the radius and tangent are perpendicular at their point of contact
∵∠OBP = ∠OAP = 90^∘
Now, In quadrilateral AOBP
∠APB + ∠AOB + ∠OBP + ∠OAP = 360^∘ [Angle sum property of a quadrilateral]
⇒ ∠APB + ∠AOB + 90^∘ + 90^∘= 360^∘
⇒ ∠APB + ∠AOB = 180^∘
Also, ∠OBP + ∠OAP = 180^∘
Since, the sum of the opposite angles of the quadrilateral is 180^∘
Hence, AOBP is a cyclic quadrilateral.

Question 12:

In two concentric circles, a chord of length 8 cm of the larger circle touches the smaller circle. If the radius of the larger circle is 5 cm then Find the radius of the smaller circle. [CBSE 2013C]

Answer:

We know that the radius and tangent are perperpendular at their point of contact
Since, the perpendicular drawn from the centre bisect the chord.
∴ AP = PB = $\frac{AB}{2}$ = 4 cm
In right triangle AOP
AO² = OP² + PA²
⇒ 5² = OP² + 4²
⇒ OP² = 9
⇒ OP = 3 cm
Hence, the radius of the smaller circle is 3 cm.

Question 13:

In the given figure, PQ is a chord of a circle with centre O and PT is a tangent. If ∠QPT = 60^∘ , find ∠PRQ

Answer:

We know that the radius and tangent are perperpendular at their point of contact.
∴∠OPT = 90^∘
Now, ∠OPQ = ∠OPT − ∠QPT = 90^∘ − 60^∘ = 30^∘
Since, OP = OQ as both are radius
∴∠OPQ = ∠OQP = 30^∘          (Angles opposite to equal sides are equal)
Now, In isosceles △POQ
∠POQ + ∠OPQ + ∠OQP = 180^∘       (Angle sum property of a triangle)
⇒ ∠POQ = 180^∘ − 30^∘ − 30^∘ = 120^∘
Now, ∠POQ + reflex ∠POQ = 360^∘         (Complete angle)
⇒ reflex ∠POQ = 360^∘ − 120^∘ = 240^∘
We know that the angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.
$∴ ∠ PRQ = \frac{1}{2} (reflex ∠ POQ) = 120 °$

Question 14:

In the given figure, PA and PB are two tangents to a circle with centre O, If ∠APB = 60^∘ then find the measure of ∠OAB

Answer:

Construction: Join OB

We know that the radius and tangent are perperpendular at their point of contact
∵∠OBP = ∠OAP = 90^∘
Now, In quadrilateral AOBP
∠AOB + ∠OBP + ∠APB + ∠OAP = 360^∘            [Angle sum property of a quadrilateral]
⇒ ∠AOB + 90^∘ + 60^∘ + 90^∘ = 360^∘
⇒ 240^∘ + ∠AOB = 360^∘
⇒ ∠AOB = 120⁰
Now, In isoceles triangle AOB
∠AOB + ∠OAB + ∠OBA = 180^∘            [Angle sum property of a triangle]
⇒ 120^∘ + 2∠OAB = 180^∘                          [∵∠OAB = ∠OBA ]
⇒ ∠OAB = 30^∘

Question 2:

A point P is 25 cm away from the centre of a circle and the length of tangent drawn from P to the circle is 24 cm. Find the radius of the circle.

Answer:

$Draw a circle and let P be a point such that OP = 25 cm . Let TP be the tangent, so that TP = 24 cm Join OT,where OT is radius . Now, tangent drawn from an external point is perpendicular to the radius at the point of contact . ∴ OT ⊥ PT In the right △ OTP, we have: {OP}^{2} = {OT}^{2} + {TP}^{2} [By Pythagoras’ theorem:] {OT}^{2} = \sqrt{{OP}^{2} - {TP}^{2}} = \sqrt{25^{2} - 24^{2}} = \sqrt{625 - 576} = \sqrt{49} =7 cm ∴ The length of the radius is 7 cm .$

Question 3:

Two concentric circles are of radii 6.5 cm and 2.5 cm. Find the length of the chord of the larger circle which touches the smaller circle

Answer:

We know that the radius and tangent are perperpendular at their point of contact
In right triangle AOP
AO² = OP² + PA²
⇒ (6.5)² = (2.5)² + PA²
⇒ PA² = 36
⇒ PA = 6 cm
Since, the perpendicular drawn from the centre bisect the chord.
∴ PA = PB = 6 cm
Now, AB = AP + PB = 6 + 6 = 12 cm
Hence, the length of the chord of the larger circle is 12 cm.

Question 4:

In the given figure, a circle inscribed in a triangle ABC, touches the sides AB, BC and AC at point D, E and F respectively. If AB = 12 cm, BC = 8 cm and AC = 10 cm, find the lengths of AD, BE and CF [CBSE 2013]

Answer:

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AD = AF, BD = BE and CE = CF
Now, AD + BD = 12 cm          …..(1)
AF + FC = 10 cm
⇒ AD + FC = 10 cm                    …..(2)
BE + EC = 8 cm
⇒ BD + FC = 8 cm                   …..(3)
Adding all these we get
AD + BD + AD + FC + BD + FC = 30
⇒2(AD + BD + FC) = 30
⇒AD + BD + FC = 15 cm           …..(4)
Solving (1) and (4), we get
FC = 3 cm
Solving (2) and (4), we get
BD = 5 cm
Solving (3) and (4), we get
and AD = 7 cm
∴ AD = AF = 7 cm, BD = BE = 5 cm and CE = CF = 3 cm

Question 5:

In the given figure, PA and PB are the tangents to a circle with centre O. Show that the points A, O, B, P are concyclic.

Answer:

$Here, O A = O B A nd O A ⊥ A P, O A ⊥ B P (S ince tangents drawn from an external point are perpendicular to the radius at the point of contact) ∴ ∠ O A P = 90^{0}, ∠ O B P = 90^{0} ∴ ∠ O A P + ∠ O B P = 90^{0} + 90^{0} = 180^{0} ∴ ∠ A O B + ∠ A P B = 180^{0} (S ince, ∠ O A P + ∠ O B P + ∠ A O B + ∠ A P B = 360^{0}) S um of opposite angle of a quadrilateral is 180 ° . Hence, A, O, B and P are concyclic .$

Question 6:

In the given figure, the chord AB of the larger of the two concentric circles, with centre O, touches the smaller circle at C. Prove that AC = CB.

Answer:

Construction: Join OA, OC and OB

We know that the radius and tangent are perperpendular at their point of contact
∴ ∠OCA = ∠OCB = 90^∘
Now, In △OCA and △OCB
∠OCA = ∠OCB = 90^∘
OA = OB (Radii of the larger circle)
OC = OC (Common)
By RHS congruency
△OCA ≅ △OCB
∴ CA = CB

Question 7:

From an external point P, tangents PA and PB are drawn to a circle with centre O. If CD is the tangent to the circle at a point E and PA = 14 cm, find the perimeter of ΔPCD.

Answer:

$Given, PA and PB are the tangents to a circle with centre O and CD is a tangent at E and PA = 14 cm . Tangents drawn from an external point are equal. ∴ PA = PB,CA = CE and DB = DE Perimeter of △ PCD = PC + CD + PD = (PA - CA) + (CE + DE) + (PB - DB) = (PA - C E) + (CE + DE) + (PB - DE) = (PA + PB) =2PA (∵ PA=PB) = (2 \times 14) cm =28 cm ∴ Perimeter of △ PCD = 28 cm .$

Question 8:

A circle is inscribed in ΔABC, touching AB, BC and AC at P, Q and R, respectively. If AB = 10 cm, AR = 7 cm and CR = 5 cm, find the length of BC.

Answer:

$Given, a circle inscribed in triangle ABC, such that the circle touches the sides of the triangle at P,Q,R . Tangents drawn to a circle from an external point are equal . ∴ AP = AR = 7 cm, CQ = CR = 5 cm . Now, BP = (AB - AP) = (10 - 7) = 3 cm ∴ BP = BQ = 3 cm ∴ BC = (BQ + QC) = > B C = 3 + 5 = > B C = 8 ∴ The length of BC is 8 cm .$

Question 9:

In the given figure, a circle touches all the four sides of a quadrilateral ABCD whose three sides are AB = 6 cm, BC = 7 cm and CD = 4 cm. Find AD.

Answer:

$Let the circle touch the sides of the quadrilateral AB,BC,CD and DA at P,Q,R and S respectively . Given, A B = 6 cm, B C = 7 cm and C D = 4 cm . Tangents drawn from an external point are equal. ∴ A P = A S, B P = B Q, C R = C Q and D R = D S \begin{array}{l} Now, A B + C D = (A P + B P) + (C R + D R) \\ = > A B + C D = (A S + B Q) + (C Q + D S) \\ = > A B + C D = (A S + D S) + (B Q + C Q) \\ = > A B + C D = A D + B C \\ = > A D = (A B + C D) - B C \\ = > A D = (6 + 4) - 7 \\ = > A D = 3 cm . \end{array} ∴ The length of A D is 3 cm .$

Question 10:

In the given figure, an isosceles triangle ABC with AB = AC, circumscribes a circle. Prove that the point of contact P bisects the base BC
[CBSE 2012]

Answer:

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AR = AQ, BR = BP and CP = CQ
Now, AB = AC
⇒ AR + RB = AQ + QC
⇒ AR + RB = AR + QC
⇒ RB = QC
⇒ BP = CP
Hence, P bisects BC at P.

Question 11:

In the given figure, O is the centre of two concentric circles of radii 4 cm and 6 cm respectively. PA and PB are tangents to the outer and inner circles, respectively. If PA = 10 cm, find the length of PB up to one decimal place.

Answer:

$Given, O is the centre of two concentric circles of radii O A = 6 cm and O B = 4 cm . P A and P B are the two tangents to the outer and inner circles respectively and P A = 10 cm . Now, tangent drawn from an external point is perpendicular to the radius at the point of contact . ∴ ∠ O A P = ∠ O B P = 90^{0} ∴ From right-angled △ O A P, O P^{2} = O A^{2} + P A^{2} \begin{array}{l} = > O P = \sqrt{O A^{2} + P A^{2}} \\ = > O P = \sqrt{6^{2} + 10^{2}} \\ = > O P = \sqrt{136} cm . \end{array} ∴ From right-angled △ O A P, O P^{2} = O B^{2} + P B^{2} \begin{array}{l} = > P B = \sqrt{O P^{2} - O B^{2}} \\ = > P B = \sqrt{136 - 16} \\ = > P B = \sqrt{120} cm \\ => P B = 10.9 cm . \end{array} ∴ The length of P B is 10.9 cm .$

Question 12:

In the given figure, a triangle ABC is drawn to circumscribe a circle of radius 3 cm such that the segments BD and DC into which BC is divided by the point of contact D are, of lengths 6 cm and 9 cm respectively. If the area of △ABC = 54 cm² then find the lengths of sides of AB and AC. [CBSE 2011, ’15]

Answer:

Construction: Join OA, OB, OC, OE ⊥ AB at E and OF ⊥ AC at F

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AE = AF, BD = BE = 6 cm and CD = CF = 9 cm
Now,
$Area (△ ABC) = Area (△ BOC) + Area (△ AOB) + Area (△ AOC) \Rightarrow 54 = \frac{1}{2} \times BC \times OD + \frac{1}{2} \times AB \times OE + \frac{1}{2} \times AC \times OF \Rightarrow 108 = 15 \times 3 + (6 + x) \times 3 + (9 + x) \times 3 \Rightarrow 36 = 15 + 6 + x + 9 + x$

$\Rightarrow 36 = 30 + 2 x \Rightarrow 2 x = 6 \Rightarrow x = 3 cm$
∴ AB = 6 + 3 = 9cm and AC = 9 + 3 = 12 cm

Question 13:

PQ is a chord of length 4.8 cm of a circle of radius 3 cm. The tangent at P and Q intersect at a point T as shown in the figure. Find the length of TP [CBSE 2013C]

Answer:

Let TR = y and TP = x
We know that the perpendicular drawn from the centre to the chord bisects it.
∴ PR = RQ
Now, PR + RQ = 4.8
⇒ PR + PR = 4.8
⇒ PR = 2.4
Now, in right triangle POR
By Using Pyhthagoras theorem, we have
PO² = OR² + PR²
⇒ 3² = OR² + (2.4)²
⇒ OR² = 3.24
⇒ OR = 1.8
Now, in right triangle TPR
By Using Pyhthagoras theorem, we have
TP² = TR² + PR²
⇒ x² = y² + (2.4)²
⇒ x² = y² + 5.76 …..(1)
Again, in right triangle TPQ
By Using Pyhthagoras theorem, we have
TO² = TP² + PO²
⇒ (y + 1.8)² = x² + 3²
⇒ y² + 3.6y + 3.24 = x² + 9
⇒ y² + 3.6y = x² + 5.76 …..(2)
Solving (1) and (2), we get
x = 4 cm and y = 3.2 cm
∴ TP = 4 cm

Question 14:

Prove that the line joining the points of contact of two parallel tangents of a circle passes through its centre. [CBSE 2014]

Answer:

Suppose CD and AB are two parallel tangents of a circle with centre O
Construction: Draw a line parallel to CD passing through O i.e, OP
We know that the radius and tangent are perperpendular at their point of contact.
∠OQC = ∠ORA = 90^∘
Now, ∠OQC + ∠POQ = 180^∘ (co-interior angles)
⇒ ∠POQ = 180^∘ − 90^∘ = 90^∘
Similarly, Now, ∠ORA + ∠POR = 180^∘ (co-interior angles)
⇒ ∠POR = 180^∘ − 90^∘ = 90^∘
Now, ∠POR + ∠POQ = 90^∘ + 90^∘ = 180^∘
Since, ∠POR and ∠POQ are linear pair angles whose sum is 180^∘
Hence, QR is a straight line passing through centre O.

Question 15:

In the given figure, a circle with centre O, is inscribed in a quadrilateral ABCD such that it touches the side BC, AB, AD and CD at points P, Q, R and S respectively. If AB = 29 cm, AD = 23 cm, ∠B = 90^∘ and DS = 5 cm then find the radius of the circle. [CBSE 2008, 13]

Answer:

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
DS = DR, AR = AQ
Now, AD = 23 cm
⇒ AR + RD = 23
⇒ AR = 23 − RD
⇒ AR = 23 − 5 [∵ DS = DR = 5]
⇒ AR = 18 cm
Again, AB = 29 cm
⇒ AQ + QB = 29
⇒ QB = 29 − AQ
⇒ QB = 29 − 18 [∵ AR = AQ = 18]
⇒ QB = 11 cm
Since all the angles are in a quadrilateral BQOP are right angles and OP = BQ.
Hence, BQOP is a square.
We know that all the sides of square are equal.
Therefore, BQ = PO = 11 cm
Hence, the radius of the circle is 11 cm.

Question 16:

In the given figure, O is the centre of the circle and TP is the tangent to the circle from an external point T. If ∠PBT = 30^∘ , prove that
BA : AT = 2 : 1 [CBSE 2015]

Answer:

AB is the chord passing through the centre
So, AB is the diameter
Since, angle in a semi circle is a right angle
∴∠APB = 90^∘
By using alternate segment theorem
We have ∠APB = ∠PAT = 30^∘
Now, in △APB
∠BAP + ∠APB + ∠BAP = 180^∘ (Angle sum property of triangle)
⇒ ∠BAP = 180^∘ − 90^∘ − 30^∘ = 60^∘
Now, ∠BAP = ∠APT + ∠PTA (Exterior angle property)
⇒ 60^∘ = 30^∘ + ∠PTA
⇒ ∠PTA = 60^∘ − 30^∘ = 30^∘
We know that sides opposite to equal angles are equal.
∴ AP = AT
In right triangle ABP
$\sin ∠ ABP = \frac{AP}{BA} \Rightarrow \sin 30 ° = \frac{AT}{BA} \Rightarrow \frac{1}{2} = \frac{AT}{BA}$
∴ BA : AT = 2 : 1

Question 1:

In the adjoining figure, a circle touches all the four sides of a quadrilateral ABCD whose sides are AB = 6 cm, BC = 9 cm and CD = 8 cm. Find the length of AD [CBSE 2011]

Answer:

We know that when a quadrilateral circumscribes a circle then sum of opposites sides is equal to the sum of other opposite sides.
∴ AB + CD = AD + BC
⇒6 + 8 = AD + 9
⇒ AD = 5 cm

Question 2:

In the given figure, PA and PB are two tangents to the circle with centre O. If ∠APB = 50^∘ then what is the measure of ∠OAB is [CBSE 2015]

Answer:

Construction: Join OB

We know that the radius and tangent are perperpendular at their point of contact
∵∠OBP = ∠OAP = 90^∘
Now, In quadrilateral AOBP
∠AOB + ∠OBP + ∠APB + ∠OAP = 360^∘            [Angle sum property of a quadrilateral]
⇒ ∠AOB + 90^∘ + 50^∘+ 90^∘ = 360^∘
⇒ 230^∘+ ∠BOC = 360^∘
⇒ ∠AOB = 130^∘
Now, In isoceles triangle AOB
∠AOB + ∠OAB + ∠OBA = 180^∘            [Angle sum property of a triangle]
⇒ 130^∘ + 2∠OAB = 180⁰                          [∵∠OAB = ∠OBA ]
⇒ ∠OAB = 25^∘

Question 3:

In the given figure, O is the centre of a circle PT and PQ are tangents to the circle from an external point P. If ∠TPQ = 70^∘ then ∠TRQ
[CBSE 2015]

Answer:

Construction: Join OQ and OT

We know that the radius and tangent are perperpendular at their point of contact
∵∠OTP = ∠OQP = 90^∘
Now, In quadrilateral OQPT
∠QOT + ∠OTP + ∠OQP + ∠TPQ = 360^∘ [Angle sum property of a quadrilateral]
⇒ ∠QOT + 90^∘ + 90^∘ + 70^∘ = 360^∘
⇒ 250^∘ + ∠QOT = 360^∘
⇒ ∠QOT = 110^∘
We know that the angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.
$∴ ∠ TRQ = \frac{1}{2} (∠ QOT) = 55 °$

Question 4:

In the given figure, common tangents AB and CD to the two circle with centres O₁ and O₂ intersect at E. Prove that AB = CD [CBSE 2014]

Answer:

We know that tangent segments to a circle from the same external point are congruent.
So, we have
EA = EC for the circle having centre O₁
and
ED = EB for the circle having centre O₁
Now, Adding ED on both sides in EA = EC, we get
EA + ED = EC + ED
⇒EA + EB = EC + ED
⇒AB = CD

Question 5:

If PT is a tangent to a circle with centre O and PQ is a chord of the circle such that ∠QPT = 70^∘ then find the measure of ∠POQ

Answer:

We know that the radius and tangent are perperpendular at their point of contact.
∴∠OPT = 90^∘
Now, ∠OPQ = ∠OPT − ∠TPQ = 90^∘ − 70^∘ = 20^∘
Since, OP = OQ as both are radius
∴∠OPQ = ∠OQP = 20^∘ (Angles opposite to equal sides are equal)
Now, In isosceles △POQ
∠POQ + ∠OPQ + ∠OQP = 180^∘ (Angle sum property of a triangle)
⇒ ∠POQ = 180^∘ − 20^∘ − 20^∘ = 140^∘

Question 6:

In the given figure, a triangle ABC is drawn to circumscribe a circle of radius 2 cm such that the segments BD and DC into which BC is divided by the point of contact D are, of lengths 4 cm and 3 cm respectively. If the area of △ABC = 21 cm² then find the lengths of sides of AB and AC. [CBSE 2011]

Answer:

Construction: Join OA, OB, OC, OE ⊥ AB at E and OF ⊥ AC at F

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AE = AF, BD = BE = 4 cm and CD = CF = 3 cm
Now,
$Area (△ ABC) = Area (△ BOC) + Area (△ AOB) + Area (△ AOC) \Rightarrow 21 = \frac{1}{2} \times BC \times OD + \frac{1}{2} \times AB \times OE + \frac{1}{2} \times AC \times OF \Rightarrow 42 = 7 \times 2 + (4 + x) \times 2 + (3 + x) \times 2 \Rightarrow 21 = 7 + 4 + x + 3 + x$

$\Rightarrow 21 = 14 + 2 x \Rightarrow 2 x = 7 \Rightarrow x = 3.5 cm$
∴ AB = 4 + 3.5 = 7.5 cm and AC = 3 + 3.5 = 6.5 cm

Question 7:

Two concentric circles are of radii 5 cm and 3 cm, respectively. Find the length of the chord of the larger circle that touches the smaller circle.

Answer:

Given: Two circles have the same centre O and AB is a chord of the larger circle touching the
smaller circle at C; also, OA=5 cm and OC=3 cm.
$In ∆ O A C, O A^{2} = O C^{2} + A C^{2} ∴ A C^{2} = O A^{2} - O C^{2} \Rightarrow A C^{2} = 5^{2} - 3^{2} \Rightarrow A C^{2} = 25 - 9 \Rightarrow A C^{2} = 16 \Rightarrow A C = 4 cm ∴ A B = 2 A C (since perpendicular drawn from the centre of the circle bisects the chord) ∴ A B = 2 \times 4 = 8 cm$
The length of the chord of the larger circle is 8 cm.

Question 8:

Prove that the perpendicular at the point of contact of the tangent to a circle passes through the centre.

Answer:

Let AB be the tangent to the circle at point P with centre O.
To prove: PQ passes through the point O.
Construction: Join OP.
Through O, draw a straight line CD parallel to the tangent AB.
Proof: Suppose that PQ doesn’t passes through point O.
PQ intersect CD at R and also intersect AB at P.
AS, CD ∥ AB, PQ is the line of intersection,
∠ORP = ∠RPA (Alternate interior angles)
but also,
∠RPA = 90^∘ (OP ⊥ AB)
⇒ ∠ORP = 90^∘
∠ROP + ∠OPA = 180^∘ (Co interior angles)
⇒∠ROP + 90^∘ = 180^∘
⇒∠ROP = 90^∘
Thus, the ΔORP has 2 right angles i.e. ∠ORP and ∠ROP which is not possible.
Hence, our supposition is wrong.
∴ PQ passes through the point O.

Question 9:

In the given figure, two tangents RQ and RP are drawn from an external point R to the circle with centre O. If ∠PRQ = 120^∘ then prove that OR = PR + RQ

Answer:

Construction: Join PO and OQ
In △POR and △QOR
OP = OQ   (Radii)
RP = RQ    (Tangents from the external point are congruent)
OR = OR    (Common)
By SSS congruency, △POR ≅ △QOR
∠PRO = ∠QRO    (C.P.C.T)
Now, ∠PRO + ∠QRO = ∠PRQ
⇒ 2∠PRO = 120^∘
⇒ ∠PRO = 60^∘
Now, In △POR
$\cos 60^{°} = \frac{PR}{OR} \Rightarrow \frac{1}{2} = \frac{PR}{OR} \Rightarrow OR = 2 PR \Rightarrow OR = PR + PR \Rightarrow OR = PR + RQ$

Question 10:

In the given figure, a circle inscribed in a triangle ABC, touches the sides AB, BC and AC at point D, E and F respectively. If AB = 14 cm, BC = 8 cm and AC = 12 cm. Find the lengths of AD, BE and CF [CBSE 2013]

Answer:

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AD = AF, BD = BE and CE = CF
Now, AD + BD = 14 cm          …..(1)
AF + FC = 12 cm
⇒ AD + FC = 12 cm                    …..(2)
BE + EC = 8 cm
⇒ BD + FC = 8 cm                   …..(3)
Adding all these we get
AD + BD + AD + FC + BD + FC = 34
⇒2(AD + BD + FC) = 34
⇒AD + BD + FC = 17 cm           …..(4)
Solving (1) and (4), we get
FC = 3 cm
Solving (2) and (4), we get
BD = 5 cm = BE
Solving (3) and (4), we get
and AD = 9 cm

Question 11:

In the given figure, O is the centre of the circle. PA and PB are tangents. Show that AOBP is a cyclic quadrilateral [CBSE 2014]

Answer:

We know that the radius and tangent are perpendicular at their point of contact
∵∠OBP = ∠OAP = 90^∘
Now, In quadrilateral AOBP
∠APB + ∠AOB + ∠OBP + ∠OAP = 360^∘ [Angle sum property of a quadrilateral]
⇒ ∠APB + ∠AOB + 90^∘ + 90^∘= 360^∘
⇒ ∠APB + ∠AOB = 180^∘
Also, ∠OBP + ∠OAP = 180^∘
Since, the sum of the opposite angles of the quadrilateral is 180^∘
Hence, AOBP is a cyclic quadrilateral.

Question 12:

In two concentric circles, a chord of length 8 cm of the larger circle touches the smaller circle. If the radius of the larger circle is 5 cm then Find the radius of the smaller circle. [CBSE 2013C]

Answer:

We know that the radius and tangent are perperpendular at their point of contact
Since, the perpendicular drawn from the centre bisect the chord.
∴ AP = PB = $\frac{AB}{2}$ = 4 cm
In right triangle AOP
AO² = OP² + PA²
⇒ 5² = OP² + 4²
⇒ OP² = 9
⇒ OP = 3 cm
Hence, the radius of the smaller circle is 3 cm.

Question 13:

In the given figure, PQ is a chord of a circle with centre O and PT is a tangent. If ∠QPT = 60^∘ , find ∠PRQ

Answer:

We know that the radius and tangent are perperpendular at their point of contact.
∴∠OPT = 90^∘
Now, ∠OPQ = ∠OPT − ∠QPT = 90^∘ − 60^∘ = 30^∘
Since, OP = OQ as both are radius
∴∠OPQ = ∠OQP = 30^∘          (Angles opposite to equal sides are equal)
Now, In isosceles △POQ
∠POQ + ∠OPQ + ∠OQP = 180^∘       (Angle sum property of a triangle)
⇒ ∠POQ = 180^∘ − 30^∘ − 30^∘ = 120^∘
Now, ∠POQ + reflex ∠POQ = 360^∘         (Complete angle)
⇒ reflex ∠POQ = 360^∘ − 120^∘ = 240^∘
We know that the angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.
$∴ ∠ PRQ = \frac{1}{2} (reflex ∠ POQ) = 120 °$

Question 14:

In the given figure, PA and PB are two tangents to a circle with centre O, If ∠APB = 60^∘ then find the measure of ∠OAB

Answer:

Construction: Join OB

We know that the radius and tangent are perperpendular at their point of contact
∵∠OBP = ∠OAP = 90^∘
Now, In quadrilateral AOBP
∠AOB + ∠OBP + ∠APB + ∠OAP = 360^∘            [Angle sum property of a quadrilateral]
⇒ ∠AOB + 90^∘ + 60^∘ + 90^∘ = 360^∘
⇒ 240^∘ + ∠AOB = 360^∘
⇒ ∠AOB = 120⁰
Now, In isoceles triangle AOB
∠AOB + ∠OAB + ∠OBA = 180^∘            [Angle sum property of a triangle]
⇒ 120^∘ + 2∠OAB = 180^∘                          [∵∠OAB = ∠OBA ]
⇒ ∠OAB = 30^∘

Question 2:

A point P is 25 cm away from the centre of a circle and the length of tangent drawn from P to the circle is 24 cm. Find the radius of the circle.

Answer:

$Draw a circle and let P be a point such that OP = 25 cm . Let TP be the tangent, so that TP = 24 cm Join OT,where OT is radius . Now, tangent drawn from an external point is perpendicular to the radius at the point of contact . ∴ OT ⊥ PT In the right △ OTP, we have: {OP}^{2} = {OT}^{2} + {TP}^{2} [By Pythagoras’ theorem:] {OT}^{2} = \sqrt{{OP}^{2} - {TP}^{2}} = \sqrt{25^{2} - 24^{2}} = \sqrt{625 - 576} = \sqrt{49} =7 cm ∴ The length of the radius is 7 cm .$

Question 3:

Two concentric circles are of radii 6.5 cm and 2.5 cm. Find the length of the chord of the larger circle which touches the smaller circle

Answer:

We know that the radius and tangent are perperpendular at their point of contact
In right triangle AOP
AO² = OP² + PA²
⇒ (6.5)² = (2.5)² + PA²
⇒ PA² = 36
⇒ PA = 6 cm
Since, the perpendicular drawn from the centre bisect the chord.
∴ PA = PB = 6 cm
Now, AB = AP + PB = 6 + 6 = 12 cm
Hence, the length of the chord of the larger circle is 12 cm.

Question 4:

In the given figure, a circle inscribed in a triangle ABC, touches the sides AB, BC and AC at point D, E and F respectively. If AB = 12 cm, BC = 8 cm and AC = 10 cm, find the lengths of AD, BE and CF [CBSE 2013]

Answer:

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AD = AF, BD = BE and CE = CF
Now, AD + BD = 12 cm          …..(1)
AF + FC = 10 cm
⇒ AD + FC = 10 cm                    …..(2)
BE + EC = 8 cm
⇒ BD + FC = 8 cm                   …..(3)
Adding all these we get
AD + BD + AD + FC + BD + FC = 30
⇒2(AD + BD + FC) = 30
⇒AD + BD + FC = 15 cm           …..(4)
Solving (1) and (4), we get
FC = 3 cm
Solving (2) and (4), we get
BD = 5 cm
Solving (3) and (4), we get
and AD = 7 cm
∴ AD = AF = 7 cm, BD = BE = 5 cm and CE = CF = 3 cm

Question 5:

In the given figure, PA and PB are the tangents to a circle with centre O. Show that the points A, O, B, P are concyclic.

Answer:

$Here, O A = O B A nd O A ⊥ A P, O A ⊥ B P (S ince tangents drawn from an external point are perpendicular to the radius at the point of contact) ∴ ∠ O A P = 90^{0}, ∠ O B P = 90^{0} ∴ ∠ O A P + ∠ O B P = 90^{0} + 90^{0} = 180^{0} ∴ ∠ A O B + ∠ A P B = 180^{0} (S ince, ∠ O A P + ∠ O B P + ∠ A O B + ∠ A P B = 360^{0}) S um of opposite angle of a quadrilateral is 180 ° . Hence, A, O, B and P are concyclic .$

Question 6:

In the given figure, the chord AB of the larger of the two concentric circles, with centre O, touches the smaller circle at C. Prove that AC = CB.

Answer:

Construction: Join OA, OC and OB

We know that the radius and tangent are perperpendular at their point of contact
∴ ∠OCA = ∠OCB = 90^∘
Now, In △OCA and △OCB
∠OCA = ∠OCB = 90^∘
OA = OB (Radii of the larger circle)
OC = OC (Common)
By RHS congruency
△OCA ≅ △OCB
∴ CA = CB

Question 7:

From an external point P, tangents PA and PB are drawn to a circle with centre O. If CD is the tangent to the circle at a point E and PA = 14 cm, find the perimeter of ΔPCD.

Answer:

$Given, PA and PB are the tangents to a circle with centre O and CD is a tangent at E and PA = 14 cm . Tangents drawn from an external point are equal. ∴ PA = PB,CA = CE and DB = DE Perimeter of △ PCD = PC + CD + PD = (PA - CA) + (CE + DE) + (PB - DB) = (PA - C E) + (CE + DE) + (PB - DE) = (PA + PB) =2PA (∵ PA=PB) = (2 \times 14) cm =28 cm ∴ Perimeter of △ PCD = 28 cm .$

Question 8:

A circle is inscribed in ΔABC, touching AB, BC and AC at P, Q and R, respectively. If AB = 10 cm, AR = 7 cm and CR = 5 cm, find the length of BC.

Answer:

$Given, a circle inscribed in triangle ABC, such that the circle touches the sides of the triangle at P,Q,R . Tangents drawn to a circle from an external point are equal . ∴ AP = AR = 7 cm, CQ = CR = 5 cm . Now, BP = (AB - AP) = (10 - 7) = 3 cm ∴ BP = BQ = 3 cm ∴ BC = (BQ + QC) = > B C = 3 + 5 = > B C = 8 ∴ The length of BC is 8 cm .$

Question 9:

In the given figure, a circle touches all the four sides of a quadrilateral ABCD whose three sides are AB = 6 cm, BC = 7 cm and CD = 4 cm. Find AD.

Answer:

$Let the circle touch the sides of the quadrilateral AB,BC,CD and DA at P,Q,R and S respectively . Given, A B = 6 cm, B C = 7 cm and C D = 4 cm . Tangents drawn from an external point are equal. ∴ A P = A S, B P = B Q, C R = C Q and D R = D S \begin{array}{l} Now, A B + C D = (A P + B P) + (C R + D R) \\ = > A B + C D = (A S + B Q) + (C Q + D S) \\ = > A B + C D = (A S + D S) + (B Q + C Q) \\ = > A B + C D = A D + B C \\ = > A D = (A B + C D) - B C \\ = > A D = (6 + 4) - 7 \\ = > A D = 3 cm . \end{array} ∴ The length of A D is 3 cm .$

Question 10:

In the given figure, an isosceles triangle ABC with AB = AC, circumscribes a circle. Prove that the point of contact P bisects the base BC
[CBSE 2012]

Answer:

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AR = AQ, BR = BP and CP = CQ
Now, AB = AC
⇒ AR + RB = AQ + QC
⇒ AR + RB = AR + QC
⇒ RB = QC
⇒ BP = CP
Hence, P bisects BC at P.

Question 11:

In the given figure, O is the centre of two concentric circles of radii 4 cm and 6 cm respectively. PA and PB are tangents to the outer and inner circles, respectively. If PA = 10 cm, find the length of PB up to one decimal place.

Answer:

$Given, O is the centre of two concentric circles of radii O A = 6 cm and O B = 4 cm . P A and P B are the two tangents to the outer and inner circles respectively and P A = 10 cm . Now, tangent drawn from an external point is perpendicular to the radius at the point of contact . ∴ ∠ O A P = ∠ O B P = 90^{0} ∴ From right-angled △ O A P, O P^{2} = O A^{2} + P A^{2} \begin{array}{l} = > O P = \sqrt{O A^{2} + P A^{2}} \\ = > O P = \sqrt{6^{2} + 10^{2}} \\ = > O P = \sqrt{136} cm . \end{array} ∴ From right-angled △ O A P, O P^{2} = O B^{2} + P B^{2} \begin{array}{l} = > P B = \sqrt{O P^{2} - O B^{2}} \\ = > P B = \sqrt{136 - 16} \\ = > P B = \sqrt{120} cm \\ => P B = 10.9 cm . \end{array} ∴ The length of P B is 10.9 cm .$

Question 12:

In the given figure, a triangle ABC is drawn to circumscribe a circle of radius 3 cm such that the segments BD and DC into which BC is divided by the point of contact D are, of lengths 6 cm and 9 cm respectively. If the area of △ABC = 54 cm² then find the lengths of sides of AB and AC. [CBSE 2011, ’15]

Answer:

Construction: Join OA, OB, OC, OE ⊥ AB at E and OF ⊥ AC at F

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AE = AF, BD = BE = 6 cm and CD = CF = 9 cm
Now,
$Area (△ ABC) = Area (△ BOC) + Area (△ AOB) + Area (△ AOC) \Rightarrow 54 = \frac{1}{2} \times BC \times OD + \frac{1}{2} \times AB \times OE + \frac{1}{2} \times AC \times OF \Rightarrow 108 = 15 \times 3 + (6 + x) \times 3 + (9 + x) \times 3 \Rightarrow 36 = 15 + 6 + x + 9 + x$

$\Rightarrow 36 = 30 + 2 x \Rightarrow 2 x = 6 \Rightarrow x = 3 cm$
∴ AB = 6 + 3 = 9cm and AC = 9 + 3 = 12 cm

Question 13:

PQ is a chord of length 4.8 cm of a circle of radius 3 cm. The tangent at P and Q intersect at a point T as shown in the figure. Find the length of TP [CBSE 2013C]

Answer:

Let TR = y and TP = x
We know that the perpendicular drawn from the centre to the chord bisects it.
∴ PR = RQ
Now, PR + RQ = 4.8
⇒ PR + PR = 4.8
⇒ PR = 2.4
Now, in right triangle POR
By Using Pyhthagoras theorem, we have
PO² = OR² + PR²
⇒ 3² = OR² + (2.4)²
⇒ OR² = 3.24
⇒ OR = 1.8
Now, in right triangle TPR
By Using Pyhthagoras theorem, we have
TP² = TR² + PR²
⇒ x² = y² + (2.4)²
⇒ x² = y² + 5.76 …..(1)
Again, in right triangle TPQ
By Using Pyhthagoras theorem, we have
TO² = TP² + PO²
⇒ (y + 1.8)² = x² + 3²
⇒ y² + 3.6y + 3.24 = x² + 9
⇒ y² + 3.6y = x² + 5.76 …..(2)
Solving (1) and (2), we get
x = 4 cm and y = 3.2 cm
∴ TP = 4 cm

Question 14:

Prove that the line joining the points of contact of two parallel tangents of a circle passes through its centre. [CBSE 2014]

Answer:

Suppose CD and AB are two parallel tangents of a circle with centre O
Construction: Draw a line parallel to CD passing through O i.e, OP
We know that the radius and tangent are perperpendular at their point of contact.
∠OQC = ∠ORA = 90^∘
Now, ∠OQC + ∠POQ = 180^∘ (co-interior angles)
⇒ ∠POQ = 180^∘ − 90^∘ = 90^∘
Similarly, Now, ∠ORA + ∠POR = 180^∘ (co-interior angles)
⇒ ∠POR = 180^∘ − 90^∘ = 90^∘
Now, ∠POR + ∠POQ = 90^∘ + 90^∘ = 180^∘
Since, ∠POR and ∠POQ are linear pair angles whose sum is 180^∘
Hence, QR is a straight line passing through centre O.

Question 15:

In the given figure, a circle with centre O, is inscribed in a quadrilateral ABCD such that it touches the side BC, AB, AD and CD at points P, Q, R and S respectively. If AB = 29 cm, AD = 23 cm, ∠B = 90^∘ and DS = 5 cm then find the radius of the circle. [CBSE 2008, 13]

Answer:

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
DS = DR, AR = AQ
Now, AD = 23 cm
⇒ AR + RD = 23
⇒ AR = 23 − RD
⇒ AR = 23 − 5 [∵ DS = DR = 5]
⇒ AR = 18 cm
Again, AB = 29 cm
⇒ AQ + QB = 29
⇒ QB = 29 − AQ
⇒ QB = 29 − 18 [∵ AR = AQ = 18]
⇒ QB = 11 cm
Since all the angles are in a quadrilateral BQOP are right angles and OP = BQ.
Hence, BQOP is a square.
We know that all the sides of square are equal.
Therefore, BQ = PO = 11 cm
Hence, the radius of the circle is 11 cm.

Question 16:

In the given figure, O is the centre of the circle and TP is the tangent to the circle from an external point T. If ∠PBT = 30^∘ , prove that
BA : AT = 2 : 1 [CBSE 2015]

Answer:

AB is the chord passing through the centre
So, AB is the diameter
Since, angle in a semi circle is a right angle
∴∠APB = 90^∘
By using alternate segment theorem
We have ∠APB = ∠PAT = 30^∘
Now, in △APB
∠BAP + ∠APB + ∠BAP = 180^∘ (Angle sum property of triangle)
⇒ ∠BAP = 180^∘ − 90^∘ − 30^∘ = 60^∘
Now, ∠BAP = ∠APT + ∠PTA (Exterior angle property)
⇒ 60^∘ = 30^∘ + ∠PTA
⇒ ∠PTA = 60^∘ − 30^∘ = 30^∘
We know that sides opposite to equal angles are equal.
∴ AP = AT
In right triangle ABP
$\sin ∠ ABP = \frac{AP}{BA} \Rightarrow \sin 30 ° = \frac{AT}{BA} \Rightarrow \frac{1}{2} = \frac{AT}{BA}$
∴ BA : AT = 2 : 1

Question 1:

In the adjoining figure, a circle touches all the four sides of a quadrilateral ABCD whose sides are AB = 6 cm, BC = 9 cm and CD = 8 cm. Find the length of AD [CBSE 2011]

Answer:

We know that when a quadrilateral circumscribes a circle then sum of opposites sides is equal to the sum of other opposite sides.
∴ AB + CD = AD + BC
⇒6 + 8 = AD + 9
⇒ AD = 5 cm

Question 2:

In the given figure, PA and PB are two tangents to the circle with centre O. If ∠APB = 50^∘ then what is the measure of ∠OAB is [CBSE 2015]

Answer:

Construction: Join OB

We know that the radius and tangent are perperpendular at their point of contact
∵∠OBP = ∠OAP = 90^∘
Now, In quadrilateral AOBP
∠AOB + ∠OBP + ∠APB + ∠OAP = 360^∘            [Angle sum property of a quadrilateral]
⇒ ∠AOB + 90^∘ + 50^∘+ 90^∘ = 360^∘
⇒ 230^∘+ ∠BOC = 360^∘
⇒ ∠AOB = 130^∘
Now, In isoceles triangle AOB
∠AOB + ∠OAB + ∠OBA = 180^∘            [Angle sum property of a triangle]
⇒ 130^∘ + 2∠OAB = 180⁰                          [∵∠OAB = ∠OBA ]
⇒ ∠OAB = 25^∘

Question 3:

In the given figure, O is the centre of a circle PT and PQ are tangents to the circle from an external point P. If ∠TPQ = 70^∘ then ∠TRQ
[CBSE 2015]

Answer:

Construction: Join OQ and OT

We know that the radius and tangent are perperpendular at their point of contact
∵∠OTP = ∠OQP = 90^∘
Now, In quadrilateral OQPT
∠QOT + ∠OTP + ∠OQP + ∠TPQ = 360^∘ [Angle sum property of a quadrilateral]
⇒ ∠QOT + 90^∘ + 90^∘ + 70^∘ = 360^∘
⇒ 250^∘ + ∠QOT = 360^∘
⇒ ∠QOT = 110^∘
We know that the angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.
$∴ ∠ TRQ = \frac{1}{2} (∠ QOT) = 55 °$

Question 4:

In the given figure, common tangents AB and CD to the two circle with centres O₁ and O₂ intersect at E. Prove that AB = CD [CBSE 2014]

Answer:

We know that tangent segments to a circle from the same external point are congruent.
So, we have
EA = EC for the circle having centre O₁
and
ED = EB for the circle having centre O₁
Now, Adding ED on both sides in EA = EC, we get
EA + ED = EC + ED
⇒EA + EB = EC + ED
⇒AB = CD

Question 5:

If PT is a tangent to a circle with centre O and PQ is a chord of the circle such that ∠QPT = 70^∘ then find the measure of ∠POQ

Answer:

We know that the radius and tangent are perperpendular at their point of contact.
∴∠OPT = 90^∘
Now, ∠OPQ = ∠OPT − ∠TPQ = 90^∘ − 70^∘ = 20^∘
Since, OP = OQ as both are radius
∴∠OPQ = ∠OQP = 20^∘ (Angles opposite to equal sides are equal)
Now, In isosceles △POQ
∠POQ + ∠OPQ + ∠OQP = 180^∘ (Angle sum property of a triangle)
⇒ ∠POQ = 180^∘ − 20^∘ − 20^∘ = 140^∘

Question 6:

In the given figure, a triangle ABC is drawn to circumscribe a circle of radius 2 cm such that the segments BD and DC into which BC is divided by the point of contact D are, of lengths 4 cm and 3 cm respectively. If the area of △ABC = 21 cm² then find the lengths of sides of AB and AC. [CBSE 2011]

Answer:

Construction: Join OA, OB, OC, OE ⊥ AB at E and OF ⊥ AC at F

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AE = AF, BD = BE = 4 cm and CD = CF = 3 cm
Now,
$Area (△ ABC) = Area (△ BOC) + Area (△ AOB) + Area (△ AOC) \Rightarrow 21 = \frac{1}{2} \times BC \times OD + \frac{1}{2} \times AB \times OE + \frac{1}{2} \times AC \times OF \Rightarrow 42 = 7 \times 2 + (4 + x) \times 2 + (3 + x) \times 2 \Rightarrow 21 = 7 + 4 + x + 3 + x$

$\Rightarrow 21 = 14 + 2 x \Rightarrow 2 x = 7 \Rightarrow x = 3.5 cm$
∴ AB = 4 + 3.5 = 7.5 cm and AC = 3 + 3.5 = 6.5 cm

Question 7:

Two concentric circles are of radii 5 cm and 3 cm, respectively. Find the length of the chord of the larger circle that touches the smaller circle.

Answer:

Given: Two circles have the same centre O and AB is a chord of the larger circle touching the
smaller circle at C; also, OA=5 cm and OC=3 cm.
$In ∆ O A C, O A^{2} = O C^{2} + A C^{2} ∴ A C^{2} = O A^{2} - O C^{2} \Rightarrow A C^{2} = 5^{2} - 3^{2} \Rightarrow A C^{2} = 25 - 9 \Rightarrow A C^{2} = 16 \Rightarrow A C = 4 cm ∴ A B = 2 A C (since perpendicular drawn from the centre of the circle bisects the chord) ∴ A B = 2 \times 4 = 8 cm$
The length of the chord of the larger circle is 8 cm.

Question 8:

Prove that the perpendicular at the point of contact of the tangent to a circle passes through the centre.

Answer:

Let AB be the tangent to the circle at point P with centre O.
To prove: PQ passes through the point O.
Construction: Join OP.
Through O, draw a straight line CD parallel to the tangent AB.
Proof: Suppose that PQ doesn’t passes through point O.
PQ intersect CD at R and also intersect AB at P.
AS, CD ∥ AB, PQ is the line of intersection,
∠ORP = ∠RPA (Alternate interior angles)
but also,
∠RPA = 90^∘ (OP ⊥ AB)
⇒ ∠ORP = 90^∘
∠ROP + ∠OPA = 180^∘ (Co interior angles)
⇒∠ROP + 90^∘ = 180^∘
⇒∠ROP = 90^∘
Thus, the ΔORP has 2 right angles i.e. ∠ORP and ∠ROP which is not possible.
Hence, our supposition is wrong.
∴ PQ passes through the point O.

Question 9:

In the given figure, two tangents RQ and RP are drawn from an external point R to the circle with centre O. If ∠PRQ = 120^∘ then prove that OR = PR + RQ

Answer:

Construction: Join PO and OQ
In △POR and △QOR
OP = OQ   (Radii)
RP = RQ    (Tangents from the external point are congruent)
OR = OR    (Common)
By SSS congruency, △POR ≅ △QOR
∠PRO = ∠QRO    (C.P.C.T)
Now, ∠PRO + ∠QRO = ∠PRQ
⇒ 2∠PRO = 120^∘
⇒ ∠PRO = 60^∘
Now, In △POR
$\cos 60^{°} = \frac{PR}{OR} \Rightarrow \frac{1}{2} = \frac{PR}{OR} \Rightarrow OR = 2 PR \Rightarrow OR = PR + PR \Rightarrow OR = PR + RQ$

Question 10:

In the given figure, a circle inscribed in a triangle ABC, touches the sides AB, BC and AC at point D, E and F respectively. If AB = 14 cm, BC = 8 cm and AC = 12 cm. Find the lengths of AD, BE and CF [CBSE 2013]

Answer:

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AD = AF, BD = BE and CE = CF
Now, AD + BD = 14 cm          …..(1)
AF + FC = 12 cm
⇒ AD + FC = 12 cm                    …..(2)
BE + EC = 8 cm
⇒ BD + FC = 8 cm                   …..(3)
Adding all these we get
AD + BD + AD + FC + BD + FC = 34
⇒2(AD + BD + FC) = 34
⇒AD + BD + FC = 17 cm           …..(4)
Solving (1) and (4), we get
FC = 3 cm
Solving (2) and (4), we get
BD = 5 cm = BE
Solving (3) and (4), we get
and AD = 9 cm

Question 11:

In the given figure, O is the centre of the circle. PA and PB are tangents. Show that AOBP is a cyclic quadrilateral [CBSE 2014]

Answer:

We know that the radius and tangent are perpendicular at their point of contact
∵∠OBP = ∠OAP = 90^∘
Now, In quadrilateral AOBP
∠APB + ∠AOB + ∠OBP + ∠OAP = 360^∘ [Angle sum property of a quadrilateral]
⇒ ∠APB + ∠AOB + 90^∘ + 90^∘= 360^∘
⇒ ∠APB + ∠AOB = 180^∘
Also, ∠OBP + ∠OAP = 180^∘
Since, the sum of the opposite angles of the quadrilateral is 180^∘
Hence, AOBP is a cyclic quadrilateral.

Question 12:

In two concentric circles, a chord of length 8 cm of the larger circle touches the smaller circle. If the radius of the larger circle is 5 cm then Find the radius of the smaller circle. [CBSE 2013C]

Answer:

We know that the radius and tangent are perperpendular at their point of contact
Since, the perpendicular drawn from the centre bisect the chord.
∴ AP = PB = $\frac{AB}{2}$ = 4 cm
In right triangle AOP
AO² = OP² + PA²
⇒ 5² = OP² + 4²
⇒ OP² = 9
⇒ OP = 3 cm
Hence, the radius of the smaller circle is 3 cm.

Question 13:

In the given figure, PQ is a chord of a circle with centre O and PT is a tangent. If ∠QPT = 60^∘ , find ∠PRQ

Answer:

We know that the radius and tangent are perperpendular at their point of contact.
∴∠OPT = 90^∘
Now, ∠OPQ = ∠OPT − ∠QPT = 90^∘ − 60^∘ = 30^∘
Since, OP = OQ as both are radius
∴∠OPQ = ∠OQP = 30^∘          (Angles opposite to equal sides are equal)
Now, In isosceles △POQ
∠POQ + ∠OPQ + ∠OQP = 180^∘       (Angle sum property of a triangle)
⇒ ∠POQ = 180^∘ − 30^∘ − 30^∘ = 120^∘
Now, ∠POQ + reflex ∠POQ = 360^∘         (Complete angle)
⇒ reflex ∠POQ = 360^∘ − 120^∘ = 240^∘
We know that the angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.
$∴ ∠ PRQ = \frac{1}{2} (reflex ∠ POQ) = 120 °$

Question 14:

In the given figure, PA and PB are two tangents to a circle with centre O, If ∠APB = 60^∘ then find the measure of ∠OAB

Answer:

Construction: Join OB

We know that the radius and tangent are perperpendular at their point of contact
∵∠OBP = ∠OAP = 90^∘
Now, In quadrilateral AOBP
∠AOB + ∠OBP + ∠APB + ∠OAP = 360^∘            [Angle sum property of a quadrilateral]
⇒ ∠AOB + 90^∘ + 60^∘ + 90^∘ = 360^∘
⇒ 240^∘ + ∠AOB = 360^∘
⇒ ∠AOB = 120⁰
Now, In isoceles triangle AOB
∠AOB + ∠OAB + ∠OBA = 180^∘            [Angle sum property of a triangle]
⇒ 120^∘ + 2∠OAB = 180^∘                          [∵∠OAB = ∠OBA ]
⇒ ∠OAB = 30^∘

Question 15:

If the angle between two tangents drawn from an external point P to a circle of radius a and centre O, is 60° then find the length of OP.

Answer:

Let PA and PB be the two tangents drawn to the circle with centre O and radius a such that $∠ APB = 60 ° .$

In ∆OPB and ∆OPA
OB = OA = a    (Radii of the circle)
$∠ OBP = ∠ OAP = 90 °$ (Tangents are perpendicular to radius at the point of contact)
BP = PA    (Lengths of tangents drawn from an external point to the circle are equal)
So, ∆OPB ≌ ∆OPA           (SAS Congruence Axiom)
$∴ ∠ OPB = ∠ OPA = 30 °$ (CPCT)
Now,
In ∆OPB,
$\sin 30 ° = \frac{OB}{OP} \Rightarrow \frac{1}{2} = \frac{a}{OP} \Rightarrow OP = 2 a$
Thus, the length of OP is 2a.
Disclaimer: The answer given in the book is incorrect.

Question 1:

The number of tangents that can be drawn from an external point to a circle is [CBSE 2011, 12]
(1) 1
(2) 2
(3) 3
(4) 4

Answer:

We can draw only two tangents from an external point to a circle.

Hence, the correct answer is option (b)

Question 2:

In the given figure, RQ is a tangent to the circle with centre O. If SQ = 6 cm and QR = 4 cm, then OR is equal to [CBSE 2014]
(a) 2.5 cm
(b) 3 cm
(c) 5 cm
(d) 8 cm

Answer:

We know that the radius and tangent are perperpendular at their point of contact
$OQ = \frac{1}{2} QS = 3 cm$ [∵Radius is half of diameter]
Now, in right triangle OQR
By using Pythagoras theorem, we have
OR² = RQ² + OQ²
= 4² + 3²
= 16 + 9
= 25
∴OR² = 25
⇒OR = 5 cm
Hence, the correct answer is option (c).

Question 3:

In a circle of radius 7 cm, tangent PT is drawn from a point P, such that PT = 24 cm. If O is the centre of the circle, then OP = ?

(a) 30 cm
(b) 28 cm
(c) 25 cm
(d) 18 cm

Answer:

(c) 25 cm
The tangent at any point of a circle is perpendicular to the radius at the point of contact.
$∴ O T ⊥ P T From right - angled triangle P T O, ∴ O P^{2} = O T^{2} + P T^{2} [U s i n g P y t h a g o r a s ‘ t h e o r e m] \Rightarrow O P^{2} = {(7)}^{2} + {(24)}^{2} \Rightarrow O P^{2} = 49 + 576 \Rightarrow O P^{2} = 625 \Rightarrow O P = \sqrt{625} \Rightarrow O P = 25 cm$

Question 4:

Which of the following pair of lines in a circle cannot be parallel?
(a) two chords
(b) a chord and a tangent
(c) two tangents
(d) two diameters

Answer:

Two diameters cannot be parallel as they perpendicularly bisect each other.
Hence, the correct answer is option (d)

Question 5:

The chord of a circle of radius 10 cm subtends a right angle at its centre. The length of the chord (in cm) is [CBSE 2014]
(a) $\frac{5}{\sqrt{2}}$ cm
(b) $5 \sqrt{2}$ cm
(c) $10 \sqrt{2}$ cm
(d) $10 \sqrt{3}$ cm

Answer:

In right triangle AOB
By using Pythagoras theorem, we have
AB² = BO² + OA²
= 10² + 10²
= 100 + 100
= 200
∴OR² = 200
⇒OR = $10 \sqrt{2}$ cm
Hence, the correct answer is option (c).

Question 6:

In the given figure, PT is a tangent to a circle with centre O. If OT = 6 cm and OP = 10 cm, then the length of tangent PT is
(a) 8 cm
(b) 10 cm
(c) 12 cm
(d) 16 cm

Answer:

In right triangle PTO
By using Pythagoras theorem, we have
PO² = OT² + TP²
⇒ 10² = 6² + TP²
⇒100 = 36 + TP²
⇒TP² = 64
⇒ TP = 8 cm
Hence, the correct answer is option (a).

Question 7:

In the given figure, point P is 26 cm away from the centre O of a circle and the length PT of the tangent drawn from P to the circle is 24 cm. Then the radius of the circle is [CBSE 2011, 12]
(a) 10 cm
(b) 12 cm
(c) 13 cm
(d) 15 cm

Answer:

Construction: Join OT

We know that the radius and tangent are perperpendular at their point of contact
In right triangle PTO
By using Pythagoras theorem, we have
PO² = OT² + TP²
⇒ 26² = OT² + 24²
⇒676 = OT² + 576
⇒TP² = 100
⇒ TP = 10 cm
Hence, the correct answer is option (a).

Question 8:

PQ is a tangent to a circle with centre O at the point P. If △OPQ is an isosceles triangle, then ∠OQP is equal to [CBSE 2014]
(a) 30^∘
(b) 45^∘
(c) 60^∘
(d) 90^∘

Answer:

We know that the radius and tangent are perperpendular at their point of contact
Now, In isoceles right triangle POQ
∠POQ + ∠OPQ + ∠OQP = 180^∘ [Angle sum property of a triangle]
⇒ 2∠OQP + 90^∘ = 180^∘
⇒ ∠OQP = 45^∘
Hence, the correct answer is option (b).

Question 9:

In the given figure, AB and AC are tangents to a circle with centre O such that ∠BAC = 40^∘ .Then ∠BOC is equal to [CBSE 2011, 14]
(a) 80^∘
(b) 100^∘
(c) 120^∘
(d) 140^∘

Answer:

We know that the radius and tangent are perperpendular at their point of contact
∵∠OBA = ∠OCA = 90^∘
Now, In quadrilateral ABOC
∠BAC + ∠OCA + ∠OBA + ∠BOC = 360^∘ [Angle sum property of a quadrilateral]
⇒ 40^∘ + 90^∘ + 90^∘ + ∠BOC = 360^∘
⇒ 220^∘ + ∠BOC = 360^∘
⇒ ∠BOC = 140^∘
Hence, the correct answer is option (d).

Question 10:

If a chord AB subtends an angle of 60^∘ at the centre of a circle, then the angle between the tangents to the circle drawn from A and B isl to [CBSE 2013C]
(a) 30^∘
(b) 60^∘
(c) 90^∘
(d) 120^∘

Answer:

We know that the radius and tangent are perperpendular at their point of contact
∵∠OBC = ∠OAC = 90^∘
Now, In quadrilateral ABOC
∠ACB + ∠OAC + ∠OBC + ∠AOB = 360^∘ [Angle sum property of a quadrilateral]
⇒ ∠ACB + 90^∘ + 90^∘ + 60^∘ = 360^∘
⇒ ∠ACB + 240^∘ = 360^∘
⇒ ∠ACB = 120^∘
Hence, the correct answer is option (d).

Question 11:

In the given figure, O is the centre of two concentric circles of radii 6 cm and 10 cm. AB is a chord of outer circle which touches the inner circle. The length of AB is
(a) 8 cm
(b) 14 cm
(c) 16 cm
(d) $\sqrt{136}$ cm

Answer:

We know that the radius and tangent are perperpendular at their point of contact
In right triangle AOP
AO² = OP² + PA²
⇒ 10² = 6² + PA²
⇒ PA² = 64
⇒ PA = 8 cm
Since, the perpendicular drawn from the centre bisect the chord.
∴ PA = PB = 8 cm
Now, AB = AP + PB = 8 + 8 = 16 cm
Hence, the correct answer is option (c).

Question 12:

In the given figure, AB and AC are tangents to a circle with centre O and radius 8 cm. If OA = 17 cm, then the length of AC (in cm) is
(a) 9 cm
(b) 15 cm
(c) $\sqrt{353}$ cm
(d) 25 cm

Answer:

We know that the radius and tangent are perperpendular at their point of contact
In right triangle AOB
By using Pythagoras theorem, we have
OA² = AB² + OB²
⇒ 17² = AB² + 8²
⇒289 = AB² + 64
⇒AB² = 225
⇒ AB = 15 cm
The tangents drawn from the external point are equal.
Therefore, the length of AC is 15 cm
Hence, the correct answer is option (b).

Question 13:

In the given figure, O is the centre of a circle. AOC is its diameter, such that ∠ACB = 50°. If AT is the tangent to the circle at the point A, then ∠BAT = ?

(a) 40°
(b) 50°
(c) 60°
(d) 65°

Answer:

(b) 50°
$∠ A B C = 90^{0} (A ngle in a semicircle) I n △ A B C, w e h a v e : ∠ A C B + ∠ C A B + ∠ A B C = 180^{0} \Rightarrow 50^{0} + ∠ C A B + 90^{0} = 180^{0} \Rightarrow ∠ C A B = (180^{0} - 140^{0}) \Rightarrow ∠ C A B = 40^{0} Now, ∠ C A T = 90^{0} (T angents drawn from an external point are perpendicular to the radius at the point of contact) ∴ ∠ C A B + ∠ B A T = 90^{0} \Rightarrow 40^{0} + ∠ B A T = 90^{0} \Rightarrow ∠ B A T = (90^{0} - 40^{0}) \Rightarrow ∠ B A T = 50^{0}$

Question 14:

In the given figure, O is the centre of the circle, PQ is a chord and PT is the tangent at P. If ∠POQ = 70^∘ , then ∠TPQ is equal to [CBSE 2011]
(a) 35^∘
(b) 45^∘
(c) 55^∘
(d) 70^∘

Answer:

We know that the radius and tangent are perperpendular at their point of contact
Since, OP = OQ
∵POQ is a isosceles right triangle
Now, In isoceles right triangle POQ
∠POQ + ∠OPQ + ∠OQP = 180^∘ [Angle sum property of a triangle]
⇒ 70^∘ + 2∠OPQ = 180^∘
⇒ ∠OPQ = 55^∘
Now, ∠TPQ + ∠OPQ = 90^∘
⇒ ∠TPQ = 35^∘
Hence, the correct answer is option (a).

Question 15:

In the given figure, AT is a tangent to the circle with centre O, such that OT = 4 cm and ∠OTA = 30°. Then, AT = ?

(a) 4 cm
(b) 2 cm
(c) $2 \sqrt{3} cm$
(d) $4 \sqrt{3} cm$

Answer:

(c) $2 \sqrt{3} cm$

$O A ⊥ A T So, \frac{A T}{O T} = \cos 30^{0} \Rightarrow \frac{A T}{4} = \frac{\sqrt{3}}{2} \Rightarrow A T = (\frac{\sqrt{3}}{2} \times 4) \Rightarrow A T = 2 \sqrt{3}$

Question 16:

If PA and PB are two tangents to a circle with centre O, such that ∠AOB = 110°, find ∠APB.

(a) 55°
(b) 60°
(c) 70°
(d) 90°

Answer:

(c) 70°
$Given, PA and PB are tangents to a circle with centre O, with ∠ AOB = 110^{0} . Now, we know that tangents drawn from an external point are perpendicular to the radius at the point of contact . So, ∠ OAP = 90^{0} and ∠ OBP = 90^{0} . \Rightarrow ∠ OAP + ∠ OBP = 90^{0} + 90^{0} = 180^{0}, which shows that OABP is a cyclic quadrilateral . ∴ ∠ AOB + ∠ APB = 180^{0} \Rightarrow 110^{0} + ∠ APB = 180^{0} \Rightarrow ∠ APB = 180^{0} - 110^{0} \Rightarrow ∠ APB = 70^{0}$

Question 17:

In the given figure, the length of BC is [CBSE 2012, ’14]
(a) 7 cm
(b) 10 cm
(c) 14 cm
(d) 15 cm

Answer:

We know that tangent segments to a circle from the same external point are congruent.
Therefore, we have
AF = AE = 4 cm
BF = BD = 3 cm
EC = AC − AE = 11 − 4 = 7 cm
CD = CE = 7 cm
∴ BC = BD + DC = 3 + 7 = 10 cm
Hence, the correct answer is option (b).

Question 18:

In the given figure, ∠AOD = 135^∘ then ∠BOC is equal to
(a) 25^∘
(b) 45^∘
(c) 52.5^∘
(d) 62.5^∘

Answer:

We know that the sum of angles subtended by opposite sides of a quadrilateral having a circumscribed circle is 180 degrees.
∴∠AOD + ∠BOC = 180^∘
⇒∠BOC = 180^∘ − 135^∘ = 45^∘
Hence, the correct answer is option (b).

Question 19:

In the given figure, O is the centre of a circle and PT is the tangent to the circle. If PQ is a chord, such that ∠QPT = 50° then ∠POQ = ?

(a) 100°
(b) 90°
(c) 80°
(d) 75°

Answer:

$(a) 100 ° Given, ∠ Q P T = 50^{0} and ∠ O P T = 90^{0} (T angents drawn from an external point are perpendicular to the radius at the point of contact) ∴ ∠ O P Q = (∠ O P T - ∠ Q P T) = (90^{0} - 50^{0}) = 40^{0} O P = O Q (R adius of the same circle) \Rightarrow ∠ O Q P = ∠ O P Q = 40^{0} In ∆ P O Q, ∠ P O Q + ∠ O Q P + ∠ O P Q = 180^{0} ∴ ∠ P O Q = 180^{0} - (40^{0} + 40^{0}) = 100^{0}$

Question 20:

In the given figure, PA and PB are two tangents to the circle with centre O. If ∠APB = 60^∘ then ∠OAB is [CBSE 2011]
(a) 15^∘
(b) 30^∘
(c) 60^∘
(d) 90^∘

Answer:

Construction: Join OB

We know that the radius and tangent are perperpendular at their point of contact
∵∠OBP = ∠OAP = 90^∘
Now, In quadrilateral AOBP
∠AOB + ∠OBP + ∠APB + ∠OAP = 360^∘            [Angle sum property of a quadrilateral]
⇒ ∠AOB + 90^∘ + 60^∘+ 90^∘ = 360^∘
⇒ 240^∘ + ∠AOB = 360^∘
⇒ ∠AOB = 120^∘
Now, In isoceles triangle AOB
∠AOB + ∠OAB + ∠OBA = 180^∘           [Angle sum property of a triangle]
⇒ 120^∘ + 2∠OAB = 180^∘                          [∵∠OAB = ∠OBA ]
⇒ ∠OAB = 30^∘
Hence, the correct answer is option (b).

Question 21:

If two tangents inclined at an angle of 60° are drawn to a circle of radius 3 cm, then the length of each tangent is

(a) 3 cm
(b) $\frac{3 \sqrt{3}}{2} cm$
(c) $3 \sqrt{3} cm$
(d) 6 cm

Answer:

(c) $3 \sqrt{3} cm$

$Given, PA and PB are tangents to circle with cent re O and radius 3 cm and ∠ APB = 60^{0} . T angents drawn from an external point are equal; so, PA = PB . A nd OP is the bisector of ∠ APB, which gives ∠ OPB = ∠ OPA = 30^{0} . OA ⊥ PA . S o, from right - angled ∆ OPA, we have : \frac{OA}{AP} = \tan 30^{0} \Rightarrow \frac{OA}{AP} = \frac{1}{\sqrt{3}} \Rightarrow \frac{3}{AP} = \frac{1}{\sqrt{3}} \Rightarrow AP = 3 \sqrt{3} cm$

Question 22:

In the given figure, PQ and PR are tangents to a circle with centre A. If ∠QPA = 27^∘ then ∠QAR equals [CBSE 2012]
(a) 63^∘
(b) 117^∘
(c) 126^∘
(d) 153^∘

Answer:

We know that the radius and tangent are perperpendular at their point of contact
Now, In △PQA
∠PQA + ∠QAP + ∠APQ = 180^∘            [Angle sum property of a triangle]
⇒ 90^∘ + ∠QAP + 27^∘ = 180^∘                          [∵∠OAB = ∠OBA ]
⇒ ∠QAP = 63^∘
In △PQA and △PRA
PQ = PR              (Tangents draw from same external point are equal)
QA = RA   (Radii of the circle)
AP = AP              (common)
By SSS congruency
△PQA ≅ △PRA
∠QAP = ∠RAP = 63^∘
∴∠QAR = ∠QAP + ∠RAP = 63^∘ + 63^∘ = 126^∘
Hence, the correct answer is option (c).

Question 23:

In the given figure, PA and PB are two tangents drawn from an external point P to a circle with centre C and radius 4 cm. If PA ⊥ PB
then the length of each tangent. [CBSE 2013]
(a) 3 cm
(b) 4 cm
(c) 5 cm
(d) 6 cm

Answer:

Construction: Join CA and CB

We know that the radius and tangent are perperpendular at their point of contact
∵∠CAP = ∠CBP = 90^∘
Since, in quadrilateral ACBP all the angles are right angles
∴ ACPB is a rectangle
Now, we know that the pair of opposite sides are equal in rectangle
∴ CB = AP and CA = BP
Therefore, CB = AP = 4 cm and CA = BP = 4 cm
Hence, the correct answer is option (b).

Question 24:

If PA and PB are two tangents to a circle with centre O, such that ∠APB = 80°, then ∠AOP = ?

(a) 40°
(b) 50°
(c) 60°
(d) 70°

Answer:

(b) 50&dedeg;
$Given, P A and P B are two tangents to a circle with centre O and ∠ A P B = 80^{0} . ∴ ∠ A P O = \frac{1}{2} ∠ A P B = 40^{0} [S ince they are equally inclined to the line segment joining the centre to that point] and ∠ O A P = 90^{0} [S ince tangents drawn from an external point are perpendicular to the radius at the point of contact] Now, in triangle A O P : ∠ A O P + ∠ O A P + ∠ A P O = 180^{0} \Rightarrow ∠ A O P + 90^{0} + 40^{0} = 180^{0} \Rightarrow ∠ A O P = 180^{0} - 130^{0} \Rightarrow ∠ A O P = 50^{0}$

Question 25:

In the given figure, O is the centre of the circle. AB is the tangent to the circle at the point P. If ∠APQ = 58^∘ then the measue of ∠PQB is [CBSE 2014]
(a) 32^∘
(b) 58^∘
(c) 122^∘
(d) 132^∘

Answer:

We know that a chord passing through the centre is the diameter of the circle.
∵∠QPR = 90^∘    (Angle in a semi circle is 90^∘)
By using alternate segment theorem
We have ∠APQ = ∠PRQ = 58^∘
Now, In △PQR
∠PQR + ∠PRQ + ∠QPR = 180⁰            [Angle sum property of a triangle]
⇒ ∠PQR + 58^∘ + 90⁰ = 180^∘
⇒ ∠PQR= 32^∘
Hence, the correct answer is option (a).

Question 26:

In the given figure, O is the centre of the circle. AB is the tangent to the circle at the point P. If ∠PAO = 30^∘ then ∠CPB + ∠ACP is equal to
(a) 60^∘
(b) 90^∘
(c) 120^∘
(d) 150^∘

Answer:

We know that a chord passing through the centre is the diameter of the circle.
∵∠DPC = 90^∘    (Angle in a semi circle is 90^∘)
Now, In △CDP
∠CDP + ∠DCP + ∠DPC = 180^∘            [Angle sum property of a triangle]
⇒ ∠CDP + ∠DCP + 90^∘ = 180^∘
⇒ ∠CDP + ∠DCP = 90^∘
By using alternate segment theorem
We have ∠CDP = ∠CPB
∴∠CPB + ∠ACP = 90^∘
Hence, the correct answer is option (b).

Question 27:

In the given figure, PQ is a tangent to a circle with centre O. A is the point of contact. If ∠PAB = 67^∘, then the measure of ∠AQB is
(a) 73^∘
(b) 64^∘
(c) 53^∘
(d) 44^∘

Answer:

We know that a chord passing through the centre is the diameter of the circle.
∵∠BAC = 90^∘    (Angle in a semi circle is 90^∘)
By using alternate segment theorem
We have ∠PAB = ∠ACB = 67^∘
Now, In △ABC
∠ABC + ∠ACB + ∠BAC = 180^∘            [Angle sum property of a triangle]
⇒ ∠ABC + 67^∘+ 90^∘ = 180^∘
⇒ ∠ABC= 23^∘
Now, ∠BAQ = 180^∘ − ∠PAB          [Linear pair angles]
= 180^∘ − 67^∘
= 113^∘
Now, In △ABQ
∠ABQ + ∠AQB + ∠BAQ = 180^∘            [Angle sum property of a triangle]
⇒ 23^∘ + ∠AQB + 113^∘ = 180^∘
⇒ ∠AQB = 44^∘
Hence, the correct answer is option (d).

Question 28:

In the given figure, two circles touch each other at C and AB is a tangent to both the circles. The measure of ∠ACB is
(a) 45^∘
(b) 60^∘
(c) 90^∘
(d) 120^∘

Answer:

We know that tangent segments to a circle from the same external point are congruent.
Therefore, we have
NA = NC and NC = NB
We also know that angle opposite to equal sides are equal
∴ ∠NAC = ∠NCA and ∠NBC = ∠NCB
Now, ∠ANC + ∠BNC = 180^∘           [Linear pair angles]
⇒ ∠NBC + ∠NCB + ∠NAC + ∠NCA= 180^∘      [Exterior angle property]
⇒ 2∠NCB + 2∠NCA= 180^∘
⇒ 2(∠NCB + ∠NCA) = 180^∘
⇒ ∠ACB = 90^∘
Hence, the correct answer is option (c).

Question 29:

O is the centre of a circle of radius 5 cm. At a distance of 13 cm from O, a point P is taken. From this point, two tangents PQ and PR are drawn to the circle. Then, the area of quadrilateral PQOR is

(a) 60 cm²
(b) 32.5 cm²
(c) 65 cm²
(d) 30 cm²

Answer:

(a) 60 cm²

$G i v e n, O Q = O R = 5 cm, O P = 13 cm . ∠ O Q P = ∠ O R P = 90^{0} (T angents drawn from an external point are perpendicular to the radius at the point of contact) From right - angled ∆ P O Q : P Q^{2} = (O P^{2} - O Q^{2}) \Rightarrow P Q^{2} = 13^{2} - 5^{2} \Rightarrow P Q^{2} = 169 - 25 \Rightarrow P Q^{2} = 144 \Rightarrow P Q = \sqrt{144} \Rightarrow P Q = 12 cm ∴ a r (△ O Q P) = \frac{1}{2} \times P Q \times O Q \Rightarrow a r (△ O Q P) = (\frac{1}{2} \times 12 \times 5) {cm}^{2} \Rightarrow a r (△ O Q P) = 30 {cm}^{2} S i m i l a r l y, a r (△ O R P) = 30 {cm}^{2} ∴ a r (q u a d . P Q O R) = (30 + 30) {cm}^{2} = 60 {cm}^{2}$

Question 30:

In the given figure, PQR is a tangent to the circle at Q, whose centre is O and AB is a chord parallel to PR, such that ∠BQR = 70°. Then, ∠AQB = ?

(a) 20°
(b) 35°
(c) 40°
(d) 45°

Answer:

(c) 40°

$S i n c e, A B ∥ P R, B Q is transversal . ∠ B Q R = ∠ A B Q = 70^{0} [A lternate angles] O Q ⊥ P Q R (T angents drawn from an external point are perpendicular to the radius at the point of contact) a n d A B ∥ P Q R ∴ Q L ⊥ A B; so, O L ⊥ A B ∴ O L bisects chord A B [P erpendicular drawn from the centre bisects the chord] From ∆ Q L A a n d ∆ Q L B : ∠ Q L A = ∠ Q L B = 90^{0} L A = L B (O L bisects chord A B) Q L is the common side . ∴ ∆ Q L A ≅ ∆ Q L B [By SAS congruency] ∴ ∠ Q A L = ∠ Q B L \Rightarrow ∠ Q A B = ∠ Q B A ∴ ∆ A Q B is isosceles . ∴ ∠ L Q A = ∠ L Q R ∠ L Q P = ∠ L Q R = 90^{0} ∠ L Q B = (90^{0} - 70^{0}) = 20^{0} ∴ ∠ L Q A = ∠ L Q B = 20^{0} \Rightarrow ∠ A Q B = ∠ L Q A + ∠ L Q B = 40^{0}$

Question 31:

The length of the tangent from an external point P to a circle of radius 5 cm is 10 cm. The distance of the point from the centre of the circle is
(a) 8 cm
(b) $\sqrt{104}$ cm
(c) 12 cm
(d) $\sqrt{125}$ cm

Answer:

We know that the radius and tangent are perperpendular at their point of contact
In right triangle PTO
By using Pythagoras theorem, we have
PO² = OT² + TP²
⇒ PO² = 5² + 10²
⇒ PO² = 25 + 100
⇒ PO² = 125
⇒ PO= $\sqrt{125}$ cm
Hence, the correct answer is option (d).

Question 32:

In the given figure, O is the centre of a circle, BOA is its diameter and the tangent at the point P meets BA extended at T. If ∠PBO = 30^∘ then ∠PTA = ?
(a) 60^∘
(b) 30^∘
(c) 15^∘
(d) 45^∘

Answer:

We know that a chord passing through the centre is the diameter of the circle.
∵∠BPA = 90^∘    (Angle in a semi circle is 90^∘)
By using alternate segment theorem
We have ∠APT = ∠ABP = 30^∘
Now, In △ABP
∠PBA + ∠BPA + ∠BAP = 180⁰            [Angle sum property of a triangle]
⇒ 30^∘ + 90⁰ + ∠BAP = 180^∘
⇒ ∠BAP = 60^∘
Now, ∠BAP = ∠APT + ∠PTA
⇒ 60^∘ = 30^∘ + ∠PTA
⇒ ∠PTA = 30^∘
Hence, the correct answer is option (b).

Question 33:

In the given figure, a circle touches the side DF of △EDF at H and touches ED and EF produced at K and M respectively. If EK = 9 cm then the perimeter of △EDF is
(a) 9 cm
(b) 12 cm
(c) 13.5 cm
(d) 18 cm

Answer:

We know that tangent segments to a circle from the same external point are congruent.
Therefore, we have
EK = EM = 9 cm
Now, EK + EM = 18 cm
⇒ ED + DK + EF + FM = 18 cm
⇒ ED + DH + EF + HF = 18 cm (∵DK = DH and FM = FH)
⇒ ED + DF + EF = 18 cm
⇒ Perimeter of △EDF = 18 cm
Hence, the correct answer is option (d)

Question 34:

To draw a pair of tangents to a circle, which are inclined to each other at angle of 45⁰ , we have to draw the tangents at the end points of those two radii, the angle between which is [CBSE 2011]
(a) 105⁰
(b) 135⁰
(c) 140⁰
(d) 145⁰

Answer:

Suppose PA and PB are two tangents we want to draw which inclined at an angle of 45⁰
We know that the radius and tangent are perperpendular at their point of contact
∵∠OBP = ∠OAP = 90⁰
Now, In quadrilateral AOBP
∠AOB + ∠OBP + + ∠OAP + ∠APB = 360⁰ [Angle sum property of a quadrilateral]
⇒ ∠AOB + 90⁰ + 90⁰ + 45⁰ = 360⁰
⇒ ∠AOB + 225⁰ = 360⁰
⇒ ∠AOB = 135⁰
Hence, the correct answer is option (b).

Question 35:

In the given figure, O is the centre of a circle; PQL and PRM are the tangents at the points Q and R respectively, and S is a point on the circle, such that ∠SQL = 50° and ∠SRM = 60°. Find ∠QSR.

(a) 40°
(b) 50°
(c) 60°
(d) 70°

Answer:

$(d) 70 ° P Q L i s a tangent O Q is the radius; s o, ∠ O Q L = 90^{0} . ∴ ∠ O Q S = (90^{0} - 50^{0}) = 40^{0} N o w, O Q = O S (Ra d i u s o f t h e s a m e c i r c l e) \Rightarrow ∠ O S Q = ∠ O Q S = 40^{0} S i m i l a r l y, ∠ O R S = (90^{0} - 60^{0}) = 30^{0}, A n d, O R = O S (R a d i u s o f t h e s a m e c i r c l e) \Rightarrow ∠ O S R = ∠ O R S = 30^{0} ∴ ∠ Q S R = ∠ O S Q + ∠ O S R \Rightarrow ∠ Q S R = (40^{0} + 30^{0}) \Rightarrow ∠ Q S R = 70^{0}$

Question 36:

In the given figure, a triangle PQR is drawn to circumscribe a circle of radius 6 cm such that the segments QT and TR into which QR is divided by the point of contact T are of lengths 12 cm and 9 cm respectively. If the area of △PQR = 189 cm² then the length of side PQ is [CBSE 2011]
(a) 17.5 cm
(b) 20 cm
(c) 22.5 cm
(d) 7.6 cm

Answer:

We know that tangent segments to a circle from the same external point are congruent.
Therefore, we have
PS = PU = x
QT = QS = 12 cm
RT = RU = 9 cm
Now,

$Ar (△ PQR) = Ar (△ POR) + Ar (△ QOR) + Ar (△ POQ) \Rightarrow 189 = \frac{1}{2} \times OU \times PR + \frac{1}{2} \times OT \times QR + \frac{1}{2} \times OS \times PQ \Rightarrow 378 = 6 \times (x + 9) + 6 \times (21) + 6 \times (12 + x) \Rightarrow 63 = x + 9 + 21 + x + 12 \Rightarrow 2 x = 21 \Rightarrow x = 10.5 cm$
Now, PQ = QS + SP = 12 + 10.5 = 22.5 cm
Hence, the correct answer is option (c ).

Question 37:

In the given figure, QR is a common tangent to the given circle, touching externally at the point T. The tangent at T meets QR at P. If PT = 3.8 cm then the length of QR is
(a) 1.9 cm
(b) 3.8 cm
(c) 5.7 cm
(d) 7.6 cm

Answer:

We know that tangent segments to a circle from the same external point are congruent.
Therefore, we have
PT = PQ = 3.8 cm and PT = PR = 3.8 cm
∴ QR = QP + PR = 3.8 + 3.8 = 7.6 cm
Hence, the correct answer is option (d)

Question 38:

In the given figure, quadrilateral ABCD is circumscribed, touching the circle at P, Q, R and S. If AP = 5 cm, BC = 7 cm and CS = 3 cm, AB = ?

(a) 9 cm
(b) 10 cm
(c) 12 cm
(d) 8 cm

Answer:

(a) 9 cm
$T angents drawn from an external point to a circle are equal . So, A Q = A P = 5 c m C R = C S = 3 c m a n d B R = (B C - C R) \Rightarrow B R = (7 - 3) c m \Rightarrow B R = 4 c m B Q = B R = 4 c m ∴ A B = (A Q + B Q) \Rightarrow A B = (5 + 4) c m \Rightarrow A B = 9 c m$

Question 39:

In the given figure, quadrilateral ABCD is circumscribed, touching the circle at P, Q, R and S. If AP = 6 cm, BP = 5 cm, CQ = 3 cm and DR = 4 cm, then the perimeter of quadrilateral ABCD is

(a) 18 cm
(b) 27 cm
(c) 36 cm
(d) 32 cm

Answer:

$(c) 36 cm Given, A P = 6 cm, B P = 5 cm, C Q = 3 cm and D R = 4 cm . T angents drawn from an external points to a circle are equal . So, A P = A S = 6 c m, B P = B Q = 5 c m, C Q = C R = 3 c m, D R = D S = 4 c m . ∴ A B = A P + P B = 6 + 5 = 11 c m B C = B Q + C Q = 5 + 3 = 8 c m C D = C R + D R = 3 + 4 = 7 c m A D = A S + D S = 6 + 4 = 10 c m ∴ Pe r i m e t e r o f q u a drilateral A BCD = AB + BC + CD + DA = (11 + 8 + 7 + 10) c m = 36 c m$

Question 40:

In the given figure, O is the centre of the circle AB is a chord and AT is the tangent at A. If ∠AOB = 100^∘ then ∠BAT is equal to
(a) 40^∘
(b) 50^∘
(c) 90^∘
(d) 100^∘

Answer:

Given: AO and BO are the radius of the circle
Since, AO = BO
∴ △AOB is an isosceles triangle.
Now, in △AOB
∠AOB + ∠OBA + ∠OAB = 180^∘ (Angle sum property of triangle)
⇒ 100^∘ + ∠OAB + ∠OAB = 180^∘ (∠OBA = ∠OAB)
⇒ 2∠OAB = 80^∘
⇒ ∠OAB = 40^∘
We know that the radius and tangent are perperpendular at their point of contact
∵∠OAT = 90^∘
⇒ ∠OAB + ∠BAT = 90^∘
⇒ ∠BAT = 90^∘ − 40^∘ = 50^∘
Hence, the correct answer is option (b).

Question 41:

In a right triangle ABC, right angled at B, BC = 12 cm and AB = 5 cm. The radius of the circle inscribed in the triangle is [CBSE 2014]
(a) 1 cm
(b) 2 cm
(c) 3 cm
(d) 4 cm

Answer:

In right triangle ABC
By using Pythagoras theorem we have
AC² = AB² + BC²
= 5² + 12²
= 25 + 144
= 169
∴ AC² = 169
⇒ AC = 13 cm
Now,

$Ar (△ ABC) = Ar (△ AOB) + Ar (△ BOC) + Ar (△ AOC) \Rightarrow \frac{1}{2} \times AB \times BC = \frac{1}{2} \times OP \times AB + \frac{1}{2} \times OQ \times BC + \frac{1}{2} \times OR \times AC \Rightarrow 5 \times 12 = x \times 5 + x \times 12 + x \times 13 \Rightarrow 60 = 30 x \Rightarrow x = 2 cm$
Hence, the correct answer is option (b).

Question 42:

In the given figure, a circle is inscribed in a quadrilateral ABCD touching its sides Ab, BC, CD and AD at P, Q, R and S respectively.
If the radius of the circle is 10 cm, BC = 38 cm, PB = 27 cm and AD ⊥ CD then the length of CD is [CBSE 2013]
(a) 11 cm
(b) 15 cm
(c) 20 cm
(d) 21 cm

Answer:

Construction: Join OR

We know that tangent segments to a circle from the same external point are congruent.
Therefore, we have
BP = BQ = 27 cm
CQ = CR
Now, BC = 38 cm
⇒ BQ + QC = 38
⇒ QC = 38 − 27 = 11 cm
Since, all the angles in quadrilateral DROS are right angles.
Hence, DROS is a rectangle.
We know that opposite sides of rectangle are equal
∴ OS = RD = 10 cm
Now, CD = CR + RD
= CQ + RD
= 11 + 10
= 21 cm
Hence, the correct answer is option(d)

Question 43:

In the given figure, ΔABC is right-angled at B, such that BC = 6 cm and AB = 8 cm. A circle with centre O has been inscribed in the triangle. OP ⊥ AB, OQ ⊥ BC and OR ⊥ AC.
If OP = OQ = OR = x cm, then x = ?

(a) 2 cm
(b) 2.5 cm
(c) 3 cm
(d) 3.5 cm

Answer:

(a) 2 cm
$Given, A B = 8 cm, B C = 6 cm Now, in ∆ A B C : A C^{2} = A B^{2} + B C^{2} \Rightarrow A C^{2} = (8^{2} + 6^{2}) \Rightarrow A C^{2} = (64 + 36) \Rightarrow A C^{2} = 100 \Rightarrow A C = \sqrt{100} \Rightarrow A C = 10 cm P B Q O is a square . C R = C Q (Since the length s of tangents drawn from an external point are equal) ∴ C Q = (B C - B Q) = (6 - x) cm Similarly, A R = A P = (A B - B P) = (8 - x) cm ∴ A C = (A R + C R) = [(8 - x) + (6 - x)] cm \Rightarrow 10 = (14 - 2 x) cm \Rightarrow 2 x = 4 \Rightarrow x = 2 cm ∴ The radius of the circle is 2 cm .$

Question 44:

Quadrilateral ABCD is circumscribed to a circle. If AB = 6 cm, BC = 7 cm and CD = 4 cm, then the length of AD is [CBSE 2012]

(a) 3 cm
(b) 4 cm
(c) 6 cm
(d) 7 cm

Answer:

We know that when a quadrilateral circumscribes a circle then sum of opposites sides is equal to the sum of other opposite sides.
∴ AB + DC = AD + BC
⇒6 + 4 = AD + 7
⇒ AD = 3 cm
Hence, the correct answer is option (a).

Question 45:

In the given figure, PA and PB are tangents to the given circle, such that PA = 5 cm and ∠APB = 60°. The length of chord AB is

(a) $5 \sqrt{2} cm$
(b) 5 cm
(c) $5 \sqrt{3} cm$
(d) 7.5 cm

Answer:

(b) 5 cm
The lengths of tangents drawn from a point to a circle are equal.
$So, P A = P B and therefore, ∠ P A B = ∠ P B A = x (say) . Then, in ∆ P A B : ∠ P A B + ∠ P B A + ∠ A P B = 180^{0} \Rightarrow x + x + 60^{0} = 180^{0} \Rightarrow 2 x = 180^{0} - 60^{0} \Rightarrow 2 x = 120^{0} \Rightarrow x = 60^{0} ∴ Each angle of △ P A B i s 60^{0} and therefore, it is an equilateral triangle . ∴ A B = P A = P B = 5 cm ∴ T he length of the chord A B is 5 cm .$

Question 46:

In the given figure, DE and DF are two tangents drawn from an external point D to a circle with centre A. If DE = 5 cm. and DE ⊥ DF then the radius of the circle is [CBSE 2013]
(a) 3 cm
(b) 4 cm
(c) 5 cm
(d) 6 cm

Answer:

Construction: Join AF and AE

We know that the radius and tangent are perperpendular at their point of contact
∵∠AED = ∠AFD = 90^∘
Since, in quadrilateral AEDF all the angles are right angles
∴ AEDF is a rectangle
Now, we know that the pair of opposite sides are equal in rectangle
∴ AF = DE = 5 cm
Therefore, the radius of the circle is 5 cm
Hence, the correct answer is option (c).

Question 47:

In the given figure, three circles with centres A, B, C, respectively, touch each other externally. If AB = 5 cm, BC = 7 cm and CA = 6 cm, the radius of the circle with centre A is

(a) 1.5 cm
(b) 2 cm
(c) 2.5 cm
(d) 3 cm

Answer:

(b) 2 cm

$Given, A B = 5 cm, B C = 7 cm a n d C A = 6 cm . Let, A R = A P = x cm . B Q = B P = y cm C R = C Q = z cm (Since the length of tangents drawn from an external point are equal) Then, A B = 5 cm \Rightarrow A P + P B = 5 cm \Rightarrow x + y = 5 \dots . . (i) Similarly, y + z = 7 \dots \dots \dots (i i) and z + x = 6 \dots . (i i i) Adding (i), (ii) and (iii), we get: (x + y) + (y + z) + (z + x) = 18 \Rightarrow 2 (x + y + z) = 18 \Rightarrow (x + y + z) = 9 \dots \dots . (i v) N o w, (i v) - (i i) : (x + y + z) - (y + z) = (9 - 7) \Rightarrow x = 2 ∴ T he radius of the circle with centre A is 2 cm .$

Assertion (A)	Reason (R)
In the given figure, a quad. ABCD is drawn to circumscribe a given circle, as shown. Then, AB + BC = AD + DC	In two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.