{"id":562886,"date":"2021-12-18T09:19:05","date_gmt":"2021-12-18T09:19:05","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=562886"},"modified":"2021-12-28T05:39:21","modified_gmt":"2021-12-28T05:39:21","slug":"maharashtra-board-solutions-class-11-arts-science-maths-part-1-chapter-6-circle","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-11-arts-science-maths-part-1-chapter-6-circle\/","title":{"rendered":"Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 1): Chapter 6- Circle"},"content":{"rendered":"\n

Class 11: Maths Chapter 6 solutions. Complete Class 11 Maths Chapter 6 Notes.<\/p>\n\n\n\n

Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 1): Chapter 6- Circle<\/h2>\n\n\n\n

Maharashtra Board 11th Maths Chapter 6, Class 11 Maths Chapter 6 solutions<\/p>\n\n\n\n

Ex 6.1<\/h4>\n\n\n\n

Question 1.
Find the equation of a circle with<\/strong>
(i) centre at origin and radius 4.
(ii) centre at (-3, -2) and radius 6.
(iii) centre at (2, -3) and radius 5.
(iv) centre at (-3, -3) passing through point (-3, -6).
Solution:
<\/strong>(i) The equation of a circle with centre at origin and radius \u2018r\u2019 is given by
x2<\/sup> + y2<\/sup> = r2<\/sup>
Here, r = 4
\u2234 The required equation of the circle is x2<\/sup> + y2<\/sup> = 42<\/sup> i.e., x2<\/sup> + y2<\/sup> = 16.<\/p>\n\n\n\n

(ii) The equation of a circle with centre at (h, k) and radius \u2018r\u2019 is given by
(x \u2013 h)2<\/sup> + (y \u2013 k)2<\/sup> = r2<\/sup>
Here, h = -3, k = -2 and r = 6
\u2234 The required equation of the circle is
[x \u2013 (-3)]2<\/sup> + [y \u2013 (-2)]2<\/sup> = 62<\/sup>
\u21d2 (x + 3)2<\/sup> + (y + 2)2<\/sup> = 36
\u21d2 x2<\/sup> + 6x + 9 + y2<\/sup> + 4y + 4 \u2013 36 = 0
\u21d2 x2<\/sup> + y2<\/sup> + 6x + 4y \u2013 23 = 0<\/p>\n\n\n\n

(iii) The equation of a circle with centre at (h, k) and radius \u2018r\u2019 is given by
(x \u2013 h)2<\/sup> + (y \u2013 k)2<\/sup> = r2<\/sup>
Here, h = 2, k = -3 and r = 5
The required equation of the circle is
(x \u2013 2)2<\/sup> + [y \u2013 (-3)]2<\/sup> = 52<\/sup>
\u21d2 (x \u2013 2)2<\/sup> + (y + 3)2<\/sup> = 25
\u21d2 x2<\/sup> \u2013 4x + 4 + y2<\/sup> + 6y + 9 \u2013 25 = 0
\u21d2 x2<\/sup> + y2<\/sup> \u2013 4x + 6y \u2013 12 = 0<\/p>\n\n\n\n

(iv) Centre of the circle is C (-3, -3) and it passes through the point P (-3, -6).
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

The equation of a circle with centre at (h, k) and radius \u2018r\u2019 is given by
(x \u2013 h)2<\/sup> + (y \u2013 k)2<\/sup> = r2<\/sup>
Here, h = -3, k = -3, r = 3
The required equation of the circle is
[x \u2013 (-3)]2<\/sup> + [y \u2013 (-3)]2<\/sup> = 32<\/sup>
\u21d2 (x + 3)2<\/sup> + (y + 3)2<\/sup> = 9
\u21d2 x2<\/sup> + 6x + 9 + y2<\/sup> + 6y + 9 \u2013 9 = 0
\u21d2 x2<\/sup> + y2<\/sup> + 6x + 6y + 9 = 0<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

Solution:
<\/strong>(i) Since the circle is touching the Y-axis, the radius of the circle is X-co-ordinate of the centre.<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

\u2234 r = a
The equation of a circle with centre at (h, k) and radius r is given by
(x \u2013 h)2<\/sup> + (y \u2013 k)2<\/sup> = r2<\/sup>
Here, h = a, k = b
The required equation of the circle is
\u21d2 (x \u2013 a)2<\/sup> + (y \u2013 b)2<\/sup> = a2<\/sup>
\u21d2 x2<\/sup> \u2013 2ax + a2<\/sup> + y2<\/sup> \u2013 2by + b2<\/sup> = a2<\/sup>
\u21d2 x2<\/sup> + y2<\/sup> \u2013 2ax \u2013 2by + b2<\/sup> = 0<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

\u21d2 h2<\/sup> = 16
\u21d2 h = \u00b14
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

the co-ordinates of the centre are (4, 0) or (-4, 0).
The equation of a circle with centre at (h, k) and radius r is given by
(x \u2013 h)2<\/sup> + (y \u2013 k)2<\/sup> = r2<\/sup>
Here, h = \u00b1 4, k = 0, r = 4
The required equation of the circle is
\u21d2 (x \u2013 4)2<\/sup> + (y \u2013 0)2<\/sup> = 42<\/sup> or (x + 4)2<\/sup> + (y \u2013 0)2<\/sup> = 42<\/sup>
\u21d2 x2<\/sup> \u2013 8x + 16 + y2<\/sup> = 16 or x2<\/sup> + 8x + 16 + y2<\/sup> = 16
\u21d2 x2<\/sup> + y2<\/sup> \u2013 8x = 0 or x2<\/sup> + y2<\/sup> + 8x = 0<\/p>\n\n\n\n

(iv) Centre of the circle is C (3, 1).
Let the circle touch the line 8x \u2013 15y + 25 = 0 at point M.
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

CM = radius (r)
CM = Length of perpendicular from centre C(3, 1) on the line 8x \u2013 15y + 25 = 0
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

The equation of a circle with centre at (h, k) and radius r is given by
(x \u2013 h)2<\/sup> + (y \u2013 k)2<\/sup> = r2<\/sup>
Here, h = 3, k = 1 and r = 2
The required equation of the circle is
\u21d2 (x \u2013 3)2<\/sup> + (y \u2013 1)2<\/sup> = 22<\/sup>
\u21d2 x2<\/sup> \u2013 6x + 9 + y2<\/sup> \u2013 2y + 1 = 4
\u21d2 x2<\/sup> + y2<\/sup> \u2013 6x \u2013 2y + 10 \u2013 4 = 0
\u21d2 x2<\/sup> + y2<\/sup> \u2013 6x \u2013 2y + 6 = 0<\/p>\n\n\n\n

Question 4.
Find the equation of the circle, if the equations of two diameters are 2x + y = 6 and 3x + 2y = 4 and radius is 9.
Solution:<\/strong>
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

Given equations of diameters are 2x + y = 6 and 3x + 2y = 4.
Let C (h, k) be the centre of the required circle.
Since point of intersection of diameters is the centre of the circle,
x = h, y = k
Equations of diameters become
2h + k = 6 \u2026..(i)
and 3h + 2k = 4 \u2026\u2026..(ii)
By (ii) \u2013 2 \u00d7 (i), we get
-h = -8
\u21d2 h = 8
Substituting h = 8 in (i), we get
2(8) + k = 6
\u21d2 k = 6 \u2013 16
\u21d2 k = -10
Centre of the circle is C (8, -10) and radius, r = 9
The equation of a circle with centre at (h, k) and radius r is given by
(x \u2013 h)2<\/sup> + (y \u2013 k)2<\/sup> = r2<\/sup>
Here, h = 8, k = -10
The required equation of the circle is
\u21d2 (x \u2013 8)2<\/sup> + (y + 10)2<\/sup> = 92<\/sup>
\u21d2 x2<\/sup> \u2013 16x + 64 + y2<\/sup> + 20y + 100 = 81
\u21d2 x2<\/sup> + y2<\/sup> \u2013 16x + 20y + 100 + 64 \u2013 81 = 0
\u21d2 x2<\/sup> + y2<\/sup> \u2013 16x + 20y + 83 = 0<\/p>\n\n\n\n

Question 5.
If y = 2x is a chord of the circle x2<\/sup> + y2<\/sup> \u2013 10x = 0, find the equation of the circle with this chord as diameter.
Solution:<\/strong>
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

y = 2x is the chord of the given circle.
It satisfies the equation of a given circle.
Substituting y = 2x in x2<\/sup> + y2<\/sup> \u2013 10x = 0, we get
\u21d2 x2<\/sup> + (2x)2<\/sup> \u2013 10x = 0
\u21d2 x2<\/sup> + 4x2<\/sup> \u2013 10x = 0
\u21d2 5x2<\/sup> \u2013 10x = 0
\u21d2 5x(x \u2013 2) = 0
\u21d2 x = 0 or x = 2
When x = 0, y = 2x = 2(0) = 0
\u2234 A = (0, 0)
When x = 2, y = 2x = 2 (2) = 4
\u2234 B = (2, 4)
End points of chord AB are A(0, 0) and B(2, 4).
Chord AB is the diameter of the required circle.
The equation of a circle having (x1<\/sub>, y1<\/sub>) and (x2<\/sub>, y2<\/sub>) as end points of diameter is given by
(x \u2013 x1<\/sub>) (x \u2013 x2<\/sub>) + (y \u2013 y1<\/sub>) (y \u2013 y2<\/sub>) = 0
Here, x1<\/sub> = 0, y1<\/sub> = 0, x2<\/sub> = 2, y2<\/sub> = 4
The required equation of the circle is
\u21d2 (x \u2013 0) (x \u2013 2) + (y \u2013 0) (y \u2013 4 ) = 0
\u21d2 x2<\/sup> \u2013 2x + y2<\/sup> \u2013 4y = 0
\u21d2 x2<\/sup> + y2<\/sup> \u2013 2x \u2013 4y = 0<\/p>\n\n\n\n

Question 6.
Find the equation of a circle with a radius of 4 units and touch both the co-ordinate axes having centre in the third quadrant. Solution:<\/strong>
The radius of the circle = 4 units
Since the circle touches both the co-ordinate axes and its centre is in the third quadrant,
the centre of the circle is C(-4, -4).
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

The equation of a circle with centre at (h, k) and radius r is given by (x \u2013 h)2<\/sup> + (y \u2013 k)2<\/sup> = r2<\/sup>
Here, h = -4, k = -4, r = 4
the required equation of the circle is
\u21d2 [x \u2013 (-4)]2<\/sup> + [y \u2013 (-4)]2<\/sup> = 42<\/sup>
\u21d2 (x + 4)2<\/sup> + (y + 4)2<\/sup> = 16
\u21d2 x2<\/sup> + 8x + 16 + y2<\/sup> + 8y + 16 \u2013 16 = 0
\u21d2 x2<\/sup> + y2<\/sup> + 8x + 8y + 16 = 0<\/p>\n\n\n\n

Question 7.
Find the equation of the circle passing through the origin and having intercepts 4 and -5 on the co-ordinate axes.
Solution:<\/strong>
Let the circle intersect X-axis at point A and intersect Y-axis at point B.
the co-ordinates of point A are (4, 0) and the co-ordinates of point B are (0, -5).
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

Since \u2220AOB is a right angle,
AB represents the diameter of the circle.
The equation of a circle having (x1<\/sub>, y1<\/sub>) and (x2<\/sub>, y2<\/sub>) as end points of diameter is given by
(x \u2013 x1<\/sub>) (x \u2013 x2<\/sub>) + (y \u2013 y1<\/sub>) (y \u2013 y2<\/sub>) = 0
Here, x1<\/sub> = 4, y1<\/sub> = 0, x2<\/sub> = 0, y2<\/sub> = -5
The required equation of the circle is
\u21d2 (x \u2013 4) (x \u2013 0) + (y \u2013 0) [y \u2013 (-5)] = 0
\u21d2 x(x \u2013 4) + y(y + 5) = 0
\u21d2 x2<\/sup> \u2013 4x + y2<\/sup> + 5y = 0
\u21d2 x2<\/sup> + y2<\/sup> \u2013 4x + 5y = 0<\/p>\n\n\n\n

Question 8.
Find the equation of a circle passing through the points (1, -4), (5, 2) and having its centre on line x \u2013 2y + 9 = 0.
Solution:<\/strong>
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

Let C(h, k) be the centre of the required circle which lies on the line x \u2013 2y + 9 = 0.
Equation of line becomes
h \u2013 2k + 9 = 0 \u2026..(i)
Also, the required circle passes through points A(1, -4) and B(5, 2).
CA = CB = radius
CA = CB
By distance formula,<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

\u21d2 x2<\/sup> + 6x + 9 + y2<\/sup> \u2013 6y + 9 \u2013 65 = 0
\u21d2 x2<\/sup> + y2<\/sup> + 6x \u2013 6y \u2013 47 = 0<\/p>\n\n\n\n

Ex 6.2<\/h4>\n\n\n\n

Question 1.
Find the centre and radius of each of the following circles:<\/strong>
(i) x2<\/sup> + y2<\/sup> \u2013 2x + 4y \u2013 4 = 0
(ii) x2<\/sup> + y2<\/sup> \u2013 6x \u2013 8y \u2013 24 = 0
(iii) 4x2<\/sup> + 4y2<\/sup> \u2013 24x \u2013 8y \u2013 24 = 0
Solution:
<\/strong>(i) Given equation of the circle is x2<\/sup> + y2<\/sup> \u2013 2x + 4y \u2013 4 = 0
Comparing this equation with x2<\/sup> + y2<\/sup> + 2gx + 2fy + c = 0, we get
2g = -2, 2f = 4 and c = -4
\u21d2 g = -1, f = 2 and c = -4
Centre of the circle = (-g, -f) = (1, -2)
and radius of the circle<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

(ii) Given equation of the circle is x2<\/sup> + y2<\/sup> \u2013 6x \u2013 8y \u2013 24 = 0
Comparing this equation with x2<\/sup> + y2<\/sup> + 2gx + 2fy + c = 0, we get
2g = -6, 2f = -8 and c = -24
\u21d2 g = -3, f = -4 and c = -24
Centre of the circle = (-g, -f) = (3, 4)
and radius of the circle
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

(iii) Given equation of the circle is 4x2<\/sup> + 4y2<\/sup> \u2013 24x \u2013 8y \u2013 24 = 0
Dividing throughout by 4, we get x2<\/sup> + y2<\/sup> \u2013 6x \u2013 2y \u2013 6 = 0
Comparing this equation with x2<\/sup> + y2<\/sup> + 2gx + 2fy + c = 0, we get
2g = -6, 2f = -2 and c = -6
\u21d2 g = -3, f = -1 and c = -6
Centre of the circle = (-g, -f) = (3, 1)
and radius of the circle
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

Question 2.
Show that the equation 3x2<\/sup> + 3y2<\/sup> + 12x + 18y \u2013 11 = 0 represents a circle.
Solution:<\/strong>
Given equation is 3x2<\/sup> + 3y2<\/sup> + 12x + 18y \u2013 11 = 0
Dividing throughout by 3, we get<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

The equation of a circle with centre at (h, k) and radius r is given by (x \u2013 h)2<\/sup>+ (y \u2013 k)2<\/sup> = r2<\/sup>
Here, h = 2, k = 3
The required equation of the circle is
(x \u2013 2)2<\/sup> + (y \u2013 3)2<\/sup> = 52<\/sup>
\u21d2 x2<\/sup> \u2013 4x + 4 + y2<\/sup> \u2013 6y + 9 = 25
\u21d2 x2<\/sup> + y2<\/sup> \u2013 4x \u2013 6y + 4 + 9 \u2013 25 = 0
\u21d2 x2<\/sup> + y2<\/sup> \u2013 4x \u2013 6y \u2013 12 = 0<\/p>\n\n\n\n

Question 4.
Show that the points (3, -2), (1, 0), (-1, -2) and (1, -4) are concyclic.
Solution:<\/strong>
Let the equation of the circle passing through the points (3, -2), (1, 0) and (-1, -2) be
x2<\/sup> + y2<\/sup> + 2gx + 2fy + c = 0 \u2026..(i)
For point (3, -2),
Substituting x = 3 and y = -2 in (i), we get
9 + 4 + 6g \u2013 4f + c = 0
\u21d2 6g \u2013 4f + c = -13 \u2026.(ii)
For point (1, 0),
Substituting x = 1 andy = 0 in (i), we get
1 + 0 + 2g + 0 + c = 0
\u21d2 2g + c = -1 \u2026\u2026(iii)
For point (-1, -2),
Substituting x = -1 and y = -2, we get
1 + 4 \u2013 2g \u2013 4f + c = 0
\u21d2 2g + 4f \u2013 c = 5 \u2026\u2026.(iv)
Adding (ii) and (iv), we get
8g = -8
\u21d2 g = -1
Substituting g = -1 in (iii), we get
-2 + c = -1
\u21d2 c = 1
Substituting g = -1 and c = 1 in (iv), we get
-2 + 4f \u2013 1 = 5
\u21d2 4f = 8
\u21d2 f = 2
Substituting g = -1, f = 2 and c = 1 in (i), we get
x2<\/sup> + y2<\/sup> \u2013 2x + 4y + 1 = 0 \u2026\u2026\u2026.(v)
If (1, -4) satisfies equation (v), the four points are concyclic.
Substituting x = 1, y = -4 in L.H.S of (v), we get
L.H.S. = (1)2<\/sup> + (-4)2<\/sup> \u2013 2(1) + 4(-4) + 1
= 1 + 16 \u2013 2 \u2013 16 + 1
= 0
= R.H.S.
Point (1, -4) satisfies equation (v).
\u2234 The given points are concyclic.<\/p>\n\n\n\n

Ex 6.3<\/h4>\n\n\n\n

Question 1.
Write the parametric equations of the circles:<\/strong>
(i) x2<\/sup> + y2<\/sup> = 9
(ii) x2<\/sup> + y2<\/sup> + 2x \u2013 4y \u2013 4 = 0
(iii) (x \u2013 3)2<\/sup> + (y + 4)2<\/sup> = 25
Solution:
<\/strong>(i) Given equation of the circle is
x2<\/sup> + y2<\/sup> = 9
\u21d2 x2<\/sup> + y2<\/sup> = 32<\/sup>
Comparing this equation with x2<\/sup> + y2<\/sup> = r2<\/sup>, we get r = 3
The parametric equations of the circle in terms of \u03b8 are
x = r cos \u03b8 and y = r sin \u03b8
\u21d2 x = 3 cos \u03b8 and y = 3 sin \u03b8<\/p>\n\n\n\n

(ii) Given equation of the circle is
x2<\/sup> + y2<\/sup> + 2x \u2013 4y \u2013 4 = 0
\u21d2 x2<\/sup> + 2x + y2<\/sup> \u2013 4y \u2013 4 = 0
\u21d2 x2<\/sup> + 2x + 1 \u2013 1 + y2<\/sup> \u2013 4y + 4 \u2013 4 \u2013 4 = 0
\u21d2 (x2<\/sup> + 2x + 1 ) + (y2<\/sup> \u2013 4y + 4) \u2013 9 = 0
\u21d2 (x + 1)2<\/sup> + (y \u2013 2)2<\/sup> = 9
\u21d2 (x + 1)2<\/sup> + (y \u2013 2)2<\/sup> = 32<\/sup>
Comparing this equation with (x \u2013 h)2<\/sup> + (y \u2013 k)2<\/sup> = r2<\/sup>, we get
h = -1, k = 2 and r = 3
The parametric equations of the circle in terms of \u03b8 are
x = h + r cos \u03b8 and y = k + r sin \u03b8
\u21d2 x = -1 + 3 cos \u03b8 and y = 2 + 3 sin \u03b8<\/p>\n\n\n\n

(iii) Given equation of the circle is
(x \u2013 3)2<\/sup> + (y + 4)2<\/sup> = 25
\u21d2 (x \u2013 3)2<\/sup> + (y + 4)2<\/sup> = 52<\/sup>
Comparing this equation with (x \u2013 h)2<\/sup> + (y \u2013 k)2<\/sup> = r2<\/sup>, we get
h = 3, k = -4 and r = 5
The parametric equations of the circle in terms of \u03b8 are
x = h + r cos \u03b8 and y = k + r sin \u03b8
\u21d2 x = 3 + 5 cos \u03b8 and y = -4 + 5 sin \u03b8<\/p>\n\n\n\n

Question 2.
Find the parametric representation of the circle 3x2<\/sup> + 3y2<\/sup> \u2013 4x + 6y \u2013 4 = 0.
Solution:<\/strong>
Given equation of the circle is 3x2<\/sup> + 3y2<\/sup> \u2013 4x + 6y \u2013 4 = 0
Dividing throughout by 3, we get
<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n

Comparing this equation with (x \u2013 h)2<\/sup> + (y \u2013 k)2<\/sup> = r2<\/sup>, we get<\/p>\n\n\n\n

\"Maharashtra<\/figure>\n\n\n\n
\"Maharashtra<\/figure>\n\n\n\n

Download PDF<\/strong><\/h2>\n\n\n\n

Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 1): Chapter 6- Circle<\/p>\n\n\n\n

Download PDF<\/strong>: Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 1): Chapter 6- Circle PDF<\/a><\/p>\n\n\n\n

Chapterwise Maharashtra Board Solutions Class 11 Arts & Science Maths (Part 1) :<\/strong><\/h2>\n\n\n\n