{"id":55383,"date":"2020-11-25T14:48:46","date_gmt":"2020-11-25T14:48:46","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=55383"},"modified":"2023-09-19T02:37:06","modified_gmt":"2023-09-19T02:37:06","slug":"ncert-solutions-for-chapter-10-circles","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-chapter-10-circles\/","title":{"rendered":"NCERT Solutions for 10th Class Maths: Chapter 10 – Circles"},"content":{"rendered":"\n

Class 10: Mathematics Chapter 10 solutions. Complete Class 10 Mathematics Chapter 10 Notes.<\/p>\n\n\n\n

NCERT Solutions for 10th Class Maths: Chapter 10 – Circles<\/strong><\/h2>\n\n\n\n

NCERT 10th Mathematics Chapter 10, class 10 Mathematics Chapter 10 solutions<\/p>\n\n\n\n

Page No: 209<\/p>\n\n\n\n

Exercise: 10.1<\/h4>\n\n\n\n

1. How many tangents can a circle have?<\/strong><\/p>\n\n\n\n

Answer<\/strong>
A circle can have infinite tangents.<\/p>\n\n\n\n

2.  Fill in the blanks :<\/strong><\/p>\n\n\n\n

(i) A tangent to a circle intersects it in …………… point(s).
(ii) A line intersecting a circle in two points is called a ………….
(iii) A circle can have …………… parallel tangents at the most.
(iv) The common point of a tangent to a circle and the circle is called …………<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) one
(ii) secant
(iii) two
(iv) point of contact<\/p>\n\n\n\n

3. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is :<\/strong><\/p>\n\n\n\n

(A) 12 cm
(B) 13 cm
(C) 8.5 cm 
(D) \u221a119 cm<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

The line drawn from the centre of the circle to the tangent is perpendicular to the tangent.
\u2234 OP \u22a5 PQ<\/p>\n\n\n\n

By Pythagoras theorem in \u0394OPQ,<\/p>\n\n\n\n

OQ2<\/sup> = OP2<\/sup> +PQ2<\/sup>
\u21d2 (12)2 <\/sup>= 52<\/sup> + PQ2<\/sup><\/p>\n\n\n\n

\u21d2PQ2<\/sup> = 144 – 25<\/p>\n\n\n\n

\u21d2PQ2<\/sup> = 119<\/p>\n\n\n\n

\u21d2PQ = \u221a119 cm<\/p>\n\n\n\n

(D) is the correct option.<\/p>\n\n\n\n

4. Draw a circle and two lines parallel to a given line such that one is a tangent and the<\/strong><\/p>\n\n\n\n

other, a secant to the circle.<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

AB and XY are two parallel lines where AB is the tangent to the circle at point C while XY is the secant to the circle.<\/p>\n\n\n\n

Exercise: 10.2<\/h2>\n\n\n\n

In Q.1 to 3, choose the correct option and give justification.<\/p>\n\n\n\n

1.  From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is<\/strong><\/p>\n\n\n\n

(A)  7 cm
(B) 12 cm
(C) 15 cm
(D) 24.5 cm<\/p>\n\n\n\n

Answer<\/strong>
<\/p>\n\n\n\n

The line drawn from the centre of the circle to the tangent is perpendicular to the tangent.<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

\u2234 OP \u22a5 PQ
also, \u0394OPQ is right angled.
OQ = 25 cm and PQ = 24 cm (Given)<\/p>\n\n\n\n

By Pythagoras theorem in \u0394OPQ,<\/p>\n\n\n\n

OQ2<\/sup> = OP2<\/sup> +PQ2<\/sup>
\u21d2 (25)2 <\/sup>= OP2<\/sup> + (24)2<\/sup><\/p>\n\n\n\n

\u21d2 OP2<\/sup> = 625 – 576<\/p>\n\n\n\n

\u21d2 OP2<\/sup> = 49<\/p>\n\n\n\n

\u21d2 OP = 7 cm<\/p>\n\n\n\n

The radius of the circle is option (A) 7 cm.<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 10, class 10 Mathematics Chapter 10 solutions<\/p>\n\n\n\n

2.  In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that \u2220POQ = 110\u00b0, then \u2220PTQ is equal to
<\/strong>(A) 60\u00b0
(B) 70\u00b0 
(C) 80\u00b0 
(D) 90\u00b0<\/p>\n\n\n\n

Answer<\/strong>

OP and OQ are radii of the circle to the tangents TP and TQ respectively.
\u2234 OP \u22a5 TP and,
\u2234 OQ \u22a5 TQ
\u2220OPT = \u2220OQT = 90\u00b0
In quadrilateral POQT,
Sum of all interior angles = 360\u00b0
\u2220PTQ + \u2220OPT + \u2220POQ + \u2220OQT  = 360\u00b0
\u21d2 \u2220PTQ + 90\u00b0 + 110\u00b0 + 90\u00b0  = 360\u00b0<\/p>\n\n\n\n

\u21d2 \u2220PTQ = 70\u00b0<\/p>\n\n\n\n

\u2220PTQ is equal to option (B) 70\u00b0.<\/p>\n\n\n\n

3.  If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80\u00b0, then \u2220 POA is equal to
<\/strong>(A) 50\u00b0  
(B) 60\u00b0 
(C) 70\u00b0
(D) 80\u00b0<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

OA and OB are radii of the circle to the tangents PA and PB respectively.
\u2234 OA \u22a5 PA and,
\u2234 OB \u22a5 PB
\u2220OBP = \u2220OAP = 90\u00b0<\/p>\n\n\n\n

In quadrilateral AOBP,
Sum of all interior angles = 360\u00b0
\u2220AOB + \u2220OBP + \u2220OAP + \u2220APB  = 360\u00b0
\u21d2 \u2220AOB + 90\u00b0 + 90\u00b0 + 80\u00b0  = 360\u00b0<\/p>\n\n\n\n

\u21d2 \u2220AOB = 100\u00b0<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

In \u0394OPB and \u0394OPA,
AP = BP (Tangents from a point are equal)
OA = OB (Radii of the circle)
OP = OP (Common side)
\u2234 \u0394OPB \u2245 \u0394OPA (by SSS congruence condition)<\/p>\n\n\n\n

Thus \u2220POB = \u2220POA<\/p>\n\n\n\n

\u2220AOB = \u2220POB + \u2220POA<\/p>\n\n\n\n

\u21d2 2 \u2220POA = \u2220AOB<\/p>\n\n\n\n

\u21d2 \u2220POA = 100\u00b0\/2 = 50\u00b0<\/p>\n\n\n\n

\u2220POA is equal to option  (A) 50\u00b0<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 10, class 10 Mathematics Chapter 10 solutions<\/p>\n\n\n\n

4.  Prove that the tangents drawn at the ends of a diameter of a circle are parallel.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Let AB be a diameter of the circle. Two tangents PQ and RS are drawn at points A and B respectively.<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Radii of the circle to the tangents will be perpendicular to it.
\u2234 OB \u22a5 RS and,<\/p>\n\n\n\n

\u2234 OA \u22a5 PQ
\u2220OBR = \u2220OBS = \u2220OAP = \u2220OAQ = 90\u00ba<\/p>\n\n\n\n

From the figure,
\u2220OBR = \u2220OAQ (Alternate interior angles)
\u2220OBS = \u2220OAP (Alternate interior angles)
Since alternate interior angles are equal, lines PQ and RS will be parallel.<\/p>\n\n\n\n

Hence Proved that the tangents drawn at the ends of a diameter of a circle are parallel.<\/p>\n\n\n\n

5.  Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Let AB be the tangent to the circle at point P with centre O.<\/p>\n\n\n\n

We have to prove that PQ passes through the point O.<\/p>\n\n\n\n

Suppose that PQ doesn’t passes through point O. Join OP.<\/p>\n\n\n\n

Through O, draw a straight line CD parallel to the tangent AB.<\/p>\n\n\n\n

PQ intersect CD at R and also intersect AB at P.<\/p>\n\n\n\n

AS, CD \/\/ AB PQ is the line of intersection,<\/p>\n\n\n\n

\u2220ORP = \u2220RPA (Alternate interior angles)<\/p>\n\n\n\n

but also,<\/p>\n\n\n\n

\u2220RPA = 90\u00b0 (PQ \u22a5 AB) <\/p>\n\n\n\n

\u21d2 \u2220ORP  = 90\u00b0<\/p>\n\n\n\n

\u2220ROP + \u2220OPA = 180\u00b0 (Co-interior angles)<\/p>\n\n\n\n

\u21d2\u2220ROP + 90\u00b0 = 180\u00b0<\/p>\n\n\n\n

\u21d2\u2220ROP = 90\u00b0<\/p>\n\n\n\n

Thus, the \u0394ORP has 2 right angles i.e. \u2220ORP  and \u2220ROP which is not possible.<\/p>\n\n\n\n

Hence, our supposition is wrong. <\/p>\n\n\n\n

\u2234 PQ passes through the point O.<\/p>\n\n\n\n

6.  The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

AB is a tangent drawn on this circle from point A.<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

\u2234 OB \u22a5 AB
OA = 5cm and AB = 4 cm (Given)
In \u0394ABO,
By Pythagoras theorem in \u0394ABO,
OA2<\/sup> =AB2 <\/sup>+ BO2<\/sup>
\u21d2 52 <\/sup>= 42 <\/sup>+ BO2<\/sup>
\u21d2 BO2<\/sup> = 25 – 16
\u21d2 BO2<\/sup> = 9
\u21d2 BO = 3
\u2234 The radius of the circle is 3 cm.<\/p>\n\n\n\n

7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Let the two concentric circles with centre O.<\/p>\n\n\n\n

AB be the chord of the larger circle which touches the smaller circle at point P. <\/p>\n\n\n\n

\u2234 AB is tangent to the smaller circle to the point P.<\/p>\n\n\n\n

\u21d2 OP \u22a5 AB
By Pythagoras theorem in \u0394OPA,
OA2<\/sup> =  AP2<\/sup> + OP2<\/sup>
\u21d2 52<\/sup> = AP2<\/sup> + 32<\/sup>
\u21d2 AP2<\/sup> = 25 – 9
\u21d2 AP = 4
In \u0394OPB,
Since OP \u22a5 AB,
AP = PB (Perpendicular from the center of the circle bisects the chord)
AB = 2AP = 2 \u00d7 4 = 8 cm
 \u2234 The length of the chord of the larger circle is 8 cm.<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 10, class 10 Mathematics Chapter 10 solutions<\/p>\n\n\n\n

8. A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD = AD + BC<\/strong><\/p>\n\n\n\n

Answer<\/strong>

From the figure we observe that,
DR = DS (Tangents on the circle from point D) \u2026 (i)
AP = AS (Tangents on the circle from point A) \u2026 (ii)<\/p>\n\n\n\n

BP = BQ (Tangents on the circle from point B) \u2026 (iii)
CR = CQ (Tangents on the circle from point C) \u2026 (iv)
Adding all these equations,
DR + AP + BP + CR = DS + AS + BQ + CQ
\u21d2 (BP + AP) + (DR + CR)  = (DS + AS) + (CQ + BQ)<\/p>\n\n\n\n

\u21d2 CD + AB = AD + BC<\/p>\n\n\n\n

9. In Fig. 10.13, XY and X\u2032Y\u2032 are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X\u2032Y\u2032 at B. Prove that \u2220 AOB = 90\u00b0.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

We joined O and C<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

A\/q,
In \u0394OPA and \u0394OCA,
OP = OC (Radii of the same circle)
AP = AC (Tangents from point A)
AO = AO (Common side)
\u2234 \u0394OPA \u2245 \u0394OCA (SSS congruence criterion)
\u21d2 \u2220POA = \u2220COA \u2026 (i)
Similarly,<\/p>\n\n\n\n

 \u0394OQB  \u2245 \u0394OCB
\u2220QOB = \u2220COB \u2026 (ii)
Since POQ is a diameter of the circle, it is a straight line.
\u2234 \u2220POA + \u2220COA + \u2220COB + \u2220QOB = 180 \u00ba
From equations (i) and (ii),
2\u2220COA + 2\u2220COB = 180\u00ba
\u21d2 \u2220COA + \u2220COB = 90\u00ba
\u21d2 \u2220AOB = 90\u00b0<\/p>\n\n\n\n

10.  Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Consider a circle with centre O. Let P be an external point from which two tangents PA and PB are drawn to the circle which are touching the circle at point A and B respectively and AB is the line segment, joining point of contacts A and B together such that it subtends \u2220AOB at center O of the circle.
It can be observed that
OA \u22a5 PA
\u2234 \u2220OAP = 90\u00b0
Similarly, OB \u22a5 PB
\u2234 \u2220OBP = 90\u00b0
In quadrilateral OAPB,
Sum of all interior angles = 360\u00ba
\u2220OAP +\u2220APB +\u2220PBO +\u2220BOA = 360\u00ba
\u21d2 90\u00ba + \u2220APB + 90\u00ba + \u2220BOA = 360\u00ba
\u21d2 \u2220APB + \u2220BOA = 180\u00ba<\/p>\n\n\n\n

\u2234 The angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 10, class 10 Mathematics Chapter 10 solutions<\/p>\n\n\n\n

11. Prove that the parallelogram circumscribing a circle is a rhombus.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

ABCD is a parallelogram,
\u2234 AB = CD … (i)
\u2234 BC = AD … (ii)<\/p>\n\n\n\n

From the figure, we observe that,<\/p>\n\n\n\n

DR = DS (Tangents to the circle at D)
CR = CQ (Tangents to the circle at C)
BP = BQ (Tangents to the circle at B)
AP = AS (Tangents to the circle at A)
Adding all these,
DR + CR + BP + AP = DS + CQ + BQ + AS
\u21d2 (DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)
\u21d2 CD + AB = AD + BC … (iii)
Putting the value of (i) and (ii) in equation (iii) we get,
2AB = 2BC
\u21d2 AB = BC … (iv)
By Comparing equations (i), (ii), and (iv) we get,
AB = BC = CD = DA
\u2234 ABCD is a rhombus.<\/p>\n\n\n\n

12. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

In \u0394ABC,<\/p>\n\n\n\n

Length of two tangents drawn from the same point to the circle are equal,
\u2234 CF = CD = 6cm
\u2234 BE = BD = 8cm
\u2234 AE = AF = x<\/em><\/p>\n\n\n\n

We observed that,
AB = AE + EB = x<\/em> + 8
BC = BD + DC = 8 + 6 = 14
CA = CF + FA = 6 + x<\/em><\/p>\n\n\n\n

Now semi perimeter of triangle (s) is,
\u21d2 2s = AB + BC + CA
= x<\/em> + 8 + 14 + 6 + x<\/em>
= 28 + 2x<\/em>
\u21d2s = 14 + x<\/em><\/p>\n\n\n\n

Area of \u0394ABC = \u221as (s – a)(s – b)(s – c)<\/p>\n\n\n\n

= \u221a(14 + x<\/em>) (14 + x <\/em>-14)(14 + x <\/em>– x<\/em> – 6)(14 + x <\/em>– x – <\/em>8)<\/p>\n\n\n\n

= \u221a(14 + x<\/em>) (x<\/em>)(8)(6)<\/p>\n\n\n\n

= \u221a(14 + x<\/em>) 48 x<\/em> … (i)<\/p>\n\n\n\n

also, Area of \u0394ABC = 2\u00d7area of (\u0394AOF + \u0394COD + \u0394DOB)<\/p>\n\n\n\n

= 2\u00d7[(1\/2\u00d7OF\u00d7AF) + (1\/2\u00d7CD\u00d7OD) + (1\/2\u00d7DB\u00d7OD)]<\/p>\n\n\n\n

= 2\u00d71\/2 (4x <\/em>+ 24 + 32) = 56 + 4x <\/em>… (ii)<\/p>\n\n\n\n

Equating equation (i) and (ii) we get,<\/p>\n\n\n\n

\u221a(14 + x<\/em>) 48 x <\/em>= 56 + 4x<\/em><\/p>\n\n\n\n

Squaring both sides,<\/p>\n\n\n\n

48x<\/em> (14 + x<\/em>) = (56 + 4x<\/em>)2<\/sup><\/p>\n\n\n\n

\u21d2 48x = <\/em>[4(14 + x)]2<\/sup>\/(14 + x<\/em>)<\/p>\n\n\n\n

\u21d2 48x = <\/em>16 (14 + x<\/em>)<\/p>\n\n\n\n

\u21d2 48x = <\/em>224 + 16x<\/em><\/p>\n\n\n\n

\u21d2 32x = <\/em>224<\/p>\n\n\n\n

\u21d2 x = <\/em>7 cm<\/p>\n\n\n\n

Hence, AB = x<\/em> + 8 = 7 + 8 = 15 cm
CA = 6 + x<\/em> = 6 + 7 = 13 cm<\/p>\n\n\n\n

NCERT 10th Mathematics Chapter 10, class 10 Mathematics Chapter 10 solutions<\/p>\n\n\n\n

13.  Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. <\/strong><\/p>\n\n\n\n

Answer<\/strong>
<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Let ABCD be a quadrilateral circumscribing a circle with O such that it touches the circle at point P, Q, R, S. Join the vertices of the quadrilateral ABCD to the center of the circle.
In \u0394OAP and \u0394OAS,
AP = AS (Tangents from the same point)
OP = OS (Radii of the circle)
OA = OA (Common side)
\u0394OAP \u2245 \u0394OAS (SSS congruence condition)
\u2234 \u2220POA = \u2220AOS<\/p>\n\n\n\n

\u21d2\u22201 = \u22208
Similarly we get,
\u22202 = \u22203
\u22204 = \u22205
\u22206 = \u22207<\/p>\n\n\n\n

Adding all these angles,
\u22201 + \u22202 + \u22203 + \u22204 + \u22205 + \u22206 + \u22207 +\u22208 = 360\u00ba
\u21d2 (\u22201 + \u22208) + (\u22202 + \u22203) + (\u22204 + \u22205) + (\u22206 + \u22207) = 360\u00ba
\u21d2 2 \u22201 + 2 \u22202 + 2 \u22205 + 2 \u22206 = 360\u00ba
\u21d2 2(\u22201 + \u22202) + 2(\u22205 + \u22206) = 360\u00ba
\u21d2 (\u22201 + \u22202) + (\u22205 + \u22206) = 180\u00ba
\u21d2 \u2220AOB + \u2220COD = 180\u00ba
Similarly, we can prove that \u2220 BOC + \u2220 DOA = 180\u00ba
Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.<\/p>\n\n\n\n

NCERT 10th Maths Chapter 10<\/p>\n\n\n\n

NCERT Solutions for 10th Class Maths: Chapter 10: Download PDF<\/strong><\/h2>\n\n\n\n

NCERT Solutions for 10th Class Maths: Chapter 10 – Circles<\/p>\n\n\n\n

Download PDF<\/strong>: NCERT Solutions for 10th Class Maths: Chapter 10 – Circles PDF<\/a><\/p>\n\n\n\n

Chapterwise NCERT Solutions for Class 10 Maths<\/strong>:<\/h2>\n\n\n\n