QBManager

Conic Section

119633 If the radius of circle $x^{2} + y^{2} - 4 x + 6 y - k = 0$ is
5 , then $k =$

1 12

2 25

3 -12

4 -25

Explanation:

A : Given,
$x^{2} + y^{2} - 4 x + 6 y - k = 0$
$x^{2} + y^{2} - 4 x + 6 y + 4 - 4 + 9 - 9 - k = 0$
$(x^{2} - 4 x + 4) + (y^{2} + 6 y + 9) - (k + 4 + 9) = 0$
$(x - 2)^{2} + (y + 3)^{2} = (k + 13)$
$r^{2} = k + 13$
$5^{2} = k + 13$
$25 = k + 13$
$k = 12$

MHT CET-2020

Conic Section

119634 The radius of the circle passing through the points $(5, 7), (2, - 2)$ and $(- 2, 0)$ is

1 2 units

2 5 units

3 3 units

4 4 units

Explanation:

B original image
Given, Point $(5, 7), (2, - 2)$ and $(- 2, 0)$ lie on circle according to figure we can say that
Radius $OA = OC$
$[h - (- 2)]^{2} + (k)^{2} = (h - 5)^{2} + (k - 7)^{2}$
$h^{2} + 4 + 4 h + k^{2} = h^{2} + 25 - 10 h + k^{2} + 49 - 14 k$
$14 h + 14 k = 70$
$h + k = 5$
Similarly, Radius $OA = OB$
$(h + 2)^{2} + (k)^{2} = (h - 2)^{2} + (k + 2)^{2}$
$h^{2} + 4 + 4 h + k^{2} = h^{2} + 4 - 4 h + k^{2} + 4 + 4 k$
$4 h + 4 h - 4 k = 4$
$8 h - 4 k = 4$
$2 h - k = 1$
$2 h = 1 + k$
$By equation (i) and (ii)$
$We get h = 2, k = 3, centre of circle (2, 3)$
$Radius of circle = OA = \sqrt{(h + 2)^{2} + (k - 0)^{2}}$
$= \sqrt{(2 + 2)^{2} + 3^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5 units$

MHT CET-2020

Conic Section

119635 If the radius of the circle
$x^{2} + y^{2} - 18 x + 12 y + k = 0$ is 11 units, then the value of $k$ is

1 -3

2 4

3 3

4 -4

Explanation:

D Given, $r = 11$ units
We know that equation of circle is
$x^{2} + y^{2} + 2 gx + 2 fy + c = 0$
$Given, equation,$
$x^{2} + y^{2} - 18 x + 12 y + k = 0$
$Comparing with both equations.$
$2 g = - 18, 2 f = + 12, c = k$
$g = - 9, f = 6, c = k$
$We also know that$
$r = \sqrt{g^{2} + f^{2} - c}$
$11 = \sqrt{81 + 36 - k}, 11 = \sqrt{117 - k}$
$121 = 117 - k$
$k = - 4$

MHT CET-2019

Conic Section

119636 The equation of the circle concentric with the circle $x^{2} + y^{2} - 6 x - 4 y - 12 = 0$ and touching the $Y$ -axis is
#[Qdiff: Hard, QCat: Numerical Based, examname: Hence, $By$ comparing with standard equation\), $2 g = - 6, 2 f = - 4, c = - 12$ , $g = - 3, f = - 2, \(∴$ centre of the circle is $C (- g, - f) = (3, 2)$ . The required circle touches the $y -$ axis then radius of the, circle is 3., the equation of the circle is-, $(x - 3)^{2} + (y - 2)^{2} = 3^{2}$ , $x^{2} + 9 - 6 x + y^{2} + 4 - 4 y = 9$ , $x^{2} + y^{2} - 6 x - 4 y + 4 = 0$ , 14. The intercept on the line $y = x$ by the circle $x^{2} + y^{2} - 2 x = 0$ is $A B$ . The equation of the circle with $AB$ as a diameter is,

1

x^{2} + y^{2} - 6 x - 4 y - 4 = 0

2

x^{2} + y^{2} - 6 x - 4 y - 9 = 0

3

x^{2} + y^{2} - 6 x - 4 y + 9 = 0

4

x^{2} + y^{2} - 6 x - 4 y + 4 = 0

Explanation:

D Given,
$x^{2} + y^{2} - 6 x - 4 y - 12 = 0$
$A s w e k n o w t h a t e q u a t i o n o f a c i r c l e i s$
$x^{2} + y^{2} + 2 gx + 2 fy + c = 0$
Ans: a
Exp: (A) : In this question we will use the general equation of the circle passing through the points of intersection of any circle. $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ and any line $a x + b y + d = 0$ equation of a circle passing through the point of intersection of $x^{2} + y^{2} - 2 x = 0$ and $y = x$ is $x^{2} + y^{2} - 2 x + λ (y - x) = 0$ $x^{2} + y^{2} - (2 + λ) x + λ y = 0$
$∴$ Its center is $(\frac{2 + λ}{2}, \frac{- λ}{2})$
Since $A B$ is the diameter of the required circle therefore, the centre must lie on $A B$ ,
$\frac{2 + λ}{2} = \frac{- λ}{2}$
$\Rightarrow λ = - 1$
$∴$ The equation of the required circle is $x^{2} + y^{2} - 2 x - 1 (y - x) = 0$ $x^{2} + y^{2} - 2 x - y + x = 0$
$x^{2} + y^{2} - x - y = 0$

Hence

Conic Section

119637 Find the equation of a circle which touches both the axes and the line $3 x - 4 y + 8 = 0$ and lies in the third quadrant.

1

x^{2} + y^{2} + 4 x - 4 y - 4 = 0

2

x^{2} + y^{2} + 4 x + 4 y + 4 = 0

3

x^{2} + y^{2} - 4 x - 4 y + 4 = 0

4

x^{2} + y^{2} + 4 x + 4 y - 4 = 0

Explanation:

B Given circle is in third quadrant and touches both axes
Then centre of circle $(- a, - a)$ So, equation of circle,
original image
$(x + a)^{2} + (y + a)^{2} = a^{2}$
$x^{2} + y^{2} + 2 a x + 2 a y + a^{2} = 0$
Perpendicular distance from the centre of the circle
$| \frac{- 3 a + 4 a + 8}{5} | = a$
$| a + 8 | = 5 a$
$a + 8 = 5 a$
$a = 2$
$x^{2} + y^{2} + 2 a x + 2 a y + a^{2} = 0$
$Putting a = 2 in equation (i) -$
$x^{2} + y^{2} + 4 x + 4 y + 4 = 0$

COMEDK-2020

Conic Section

119633 If the radius of circle $x^{2} + y^{2} - 4 x + 6 y - k = 0$ is
5 , then $k =$

1 12

2 25

3 -12

4 -25

Explanation:

A : Given,
$x^{2} + y^{2} - 4 x + 6 y - k = 0$
$x^{2} + y^{2} - 4 x + 6 y + 4 - 4 + 9 - 9 - k = 0$
$(x^{2} - 4 x + 4) + (y^{2} + 6 y + 9) - (k + 4 + 9) = 0$
$(x - 2)^{2} + (y + 3)^{2} = (k + 13)$
$r^{2} = k + 13$
$5^{2} = k + 13$
$25 = k + 13$
$k = 12$

MHT CET-2020

Conic Section

119634 The radius of the circle passing through the points $(5, 7), (2, - 2)$ and $(- 2, 0)$ is

1 2 units

2 5 units

3 3 units

4 4 units

Explanation:

B original image
Given, Point $(5, 7), (2, - 2)$ and $(- 2, 0)$ lie on circle according to figure we can say that
Radius $OA = OC$
$[h - (- 2)]^{2} + (k)^{2} = (h - 5)^{2} + (k - 7)^{2}$
$h^{2} + 4 + 4 h + k^{2} = h^{2} + 25 - 10 h + k^{2} + 49 - 14 k$
$14 h + 14 k = 70$
$h + k = 5$
Similarly, Radius $OA = OB$
$(h + 2)^{2} + (k)^{2} = (h - 2)^{2} + (k + 2)^{2}$
$h^{2} + 4 + 4 h + k^{2} = h^{2} + 4 - 4 h + k^{2} + 4 + 4 k$
$4 h + 4 h - 4 k = 4$
$8 h - 4 k = 4$
$2 h - k = 1$
$2 h = 1 + k$
$By equation (i) and (ii)$
$We get h = 2, k = 3, centre of circle (2, 3)$
$Radius of circle = OA = \sqrt{(h + 2)^{2} + (k - 0)^{2}}$
$= \sqrt{(2 + 2)^{2} + 3^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5 units$

MHT CET-2020

Conic Section

119635 If the radius of the circle
$x^{2} + y^{2} - 18 x + 12 y + k = 0$ is 11 units, then the value of $k$ is

1 -3

2 4

3 3

4 -4

Explanation:

D Given, $r = 11$ units
We know that equation of circle is
$x^{2} + y^{2} + 2 gx + 2 fy + c = 0$
$Given, equation,$
$x^{2} + y^{2} - 18 x + 12 y + k = 0$
$Comparing with both equations.$
$2 g = - 18, 2 f = + 12, c = k$
$g = - 9, f = 6, c = k$
$We also know that$
$r = \sqrt{g^{2} + f^{2} - c}$
$11 = \sqrt{81 + 36 - k}, 11 = \sqrt{117 - k}$
$121 = 117 - k$
$k = - 4$

MHT CET-2019

Conic Section

119636 The equation of the circle concentric with the circle $x^{2} + y^{2} - 6 x - 4 y - 12 = 0$ and touching the $Y$ -axis is
#[Qdiff: Hard, QCat: Numerical Based, examname: Hence, $By$ comparing with standard equation\), $2 g = - 6, 2 f = - 4, c = - 12$ , $g = - 3, f = - 2, \(∴$ centre of the circle is $C (- g, - f) = (3, 2)$ . The required circle touches the $y -$ axis then radius of the, circle is 3., the equation of the circle is-, $(x - 3)^{2} + (y - 2)^{2} = 3^{2}$ , $x^{2} + 9 - 6 x + y^{2} + 4 - 4 y = 9$ , $x^{2} + y^{2} - 6 x - 4 y + 4 = 0$ , 14. The intercept on the line $y = x$ by the circle $x^{2} + y^{2} - 2 x = 0$ is $A B$ . The equation of the circle with $AB$ as a diameter is,

1

x^{2} + y^{2} - 6 x - 4 y - 4 = 0

2

x^{2} + y^{2} - 6 x - 4 y - 9 = 0

3

x^{2} + y^{2} - 6 x - 4 y + 9 = 0

4

x^{2} + y^{2} - 6 x - 4 y + 4 = 0

Explanation:

D Given,
$x^{2} + y^{2} - 6 x - 4 y - 12 = 0$
$A s w e k n o w t h a t e q u a t i o n o f a c i r c l e i s$
$x^{2} + y^{2} + 2 gx + 2 fy + c = 0$
Ans: a
Exp: (A) : In this question we will use the general equation of the circle passing through the points of intersection of any circle. $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ and any line $a x + b y + d = 0$ equation of a circle passing through the point of intersection of $x^{2} + y^{2} - 2 x = 0$ and $y = x$ is $x^{2} + y^{2} - 2 x + λ (y - x) = 0$ $x^{2} + y^{2} - (2 + λ) x + λ y = 0$
$∴$ Its center is $(\frac{2 + λ}{2}, \frac{- λ}{2})$
Since $A B$ is the diameter of the required circle therefore, the centre must lie on $A B$ ,
$\frac{2 + λ}{2} = \frac{- λ}{2}$
$\Rightarrow λ = - 1$
$∴$ The equation of the required circle is $x^{2} + y^{2} - 2 x - 1 (y - x) = 0$ $x^{2} + y^{2} - 2 x - y + x = 0$
$x^{2} + y^{2} - x - y = 0$

Hence

Conic Section

119637 Find the equation of a circle which touches both the axes and the line $3 x - 4 y + 8 = 0$ and lies in the third quadrant.

1

x^{2} + y^{2} + 4 x - 4 y - 4 = 0

2

x^{2} + y^{2} + 4 x + 4 y + 4 = 0

3

x^{2} + y^{2} - 4 x - 4 y + 4 = 0

4

x^{2} + y^{2} + 4 x + 4 y - 4 = 0

Explanation:

B Given circle is in third quadrant and touches both axes
Then centre of circle $(- a, - a)$ So, equation of circle,
original image
$(x + a)^{2} + (y + a)^{2} = a^{2}$
$x^{2} + y^{2} + 2 a x + 2 a y + a^{2} = 0$
Perpendicular distance from the centre of the circle
$| \frac{- 3 a + 4 a + 8}{5} | = a$
$| a + 8 | = 5 a$
$a + 8 = 5 a$
$a = 2$
$x^{2} + y^{2} + 2 a x + 2 a y + a^{2} = 0$
$Putting a = 2 in equation (i) -$
$x^{2} + y^{2} + 4 x + 4 y + 4 = 0$

COMEDK-2020

Conic Section

119633 If the radius of circle $x^{2} + y^{2} - 4 x + 6 y - k = 0$ is
5 , then $k =$

1 12

2 25

3 -12

4 -25

Explanation:

A : Given,
$x^{2} + y^{2} - 4 x + 6 y - k = 0$
$x^{2} + y^{2} - 4 x + 6 y + 4 - 4 + 9 - 9 - k = 0$
$(x^{2} - 4 x + 4) + (y^{2} + 6 y + 9) - (k + 4 + 9) = 0$
$(x - 2)^{2} + (y + 3)^{2} = (k + 13)$
$r^{2} = k + 13$
$5^{2} = k + 13$
$25 = k + 13$
$k = 12$

MHT CET-2020

Conic Section

119634 The radius of the circle passing through the points $(5, 7), (2, - 2)$ and $(- 2, 0)$ is

1 2 units

2 5 units

3 3 units

4 4 units

Explanation:

B original image
Given, Point $(5, 7), (2, - 2)$ and $(- 2, 0)$ lie on circle according to figure we can say that
Radius $OA = OC$
$[h - (- 2)]^{2} + (k)^{2} = (h - 5)^{2} + (k - 7)^{2}$
$h^{2} + 4 + 4 h + k^{2} = h^{2} + 25 - 10 h + k^{2} + 49 - 14 k$
$14 h + 14 k = 70$
$h + k = 5$
Similarly, Radius $OA = OB$
$(h + 2)^{2} + (k)^{2} = (h - 2)^{2} + (k + 2)^{2}$
$h^{2} + 4 + 4 h + k^{2} = h^{2} + 4 - 4 h + k^{2} + 4 + 4 k$
$4 h + 4 h - 4 k = 4$
$8 h - 4 k = 4$
$2 h - k = 1$
$2 h = 1 + k$
$By equation (i) and (ii)$
$We get h = 2, k = 3, centre of circle (2, 3)$
$Radius of circle = OA = \sqrt{(h + 2)^{2} + (k - 0)^{2}}$
$= \sqrt{(2 + 2)^{2} + 3^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5 units$

MHT CET-2020

Conic Section

119635 If the radius of the circle
$x^{2} + y^{2} - 18 x + 12 y + k = 0$ is 11 units, then the value of $k$ is

1 -3

2 4

3 3

4 -4

Explanation:

D Given, $r = 11$ units
We know that equation of circle is
$x^{2} + y^{2} + 2 gx + 2 fy + c = 0$
$Given, equation,$
$x^{2} + y^{2} - 18 x + 12 y + k = 0$
$Comparing with both equations.$
$2 g = - 18, 2 f = + 12, c = k$
$g = - 9, f = 6, c = k$
$We also know that$
$r = \sqrt{g^{2} + f^{2} - c}$
$11 = \sqrt{81 + 36 - k}, 11 = \sqrt{117 - k}$
$121 = 117 - k$
$k = - 4$

MHT CET-2019

Conic Section

119636 The equation of the circle concentric with the circle $x^{2} + y^{2} - 6 x - 4 y - 12 = 0$ and touching the $Y$ -axis is
#[Qdiff: Hard, QCat: Numerical Based, examname: Hence, $By$ comparing with standard equation\), $2 g = - 6, 2 f = - 4, c = - 12$ , $g = - 3, f = - 2, \(∴$ centre of the circle is $C (- g, - f) = (3, 2)$ . The required circle touches the $y -$ axis then radius of the, circle is 3., the equation of the circle is-, $(x - 3)^{2} + (y - 2)^{2} = 3^{2}$ , $x^{2} + 9 - 6 x + y^{2} + 4 - 4 y = 9$ , $x^{2} + y^{2} - 6 x - 4 y + 4 = 0$ , 14. The intercept on the line $y = x$ by the circle $x^{2} + y^{2} - 2 x = 0$ is $A B$ . The equation of the circle with $AB$ as a diameter is,

1

x^{2} + y^{2} - 6 x - 4 y - 4 = 0

2

x^{2} + y^{2} - 6 x - 4 y - 9 = 0

3

x^{2} + y^{2} - 6 x - 4 y + 9 = 0

4

x^{2} + y^{2} - 6 x - 4 y + 4 = 0

Explanation:

D Given,
$x^{2} + y^{2} - 6 x - 4 y - 12 = 0$
$A s w e k n o w t h a t e q u a t i o n o f a c i r c l e i s$
$x^{2} + y^{2} + 2 gx + 2 fy + c = 0$
Ans: a
Exp: (A) : In this question we will use the general equation of the circle passing through the points of intersection of any circle. $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ and any line $a x + b y + d = 0$ equation of a circle passing through the point of intersection of $x^{2} + y^{2} - 2 x = 0$ and $y = x$ is $x^{2} + y^{2} - 2 x + λ (y - x) = 0$ $x^{2} + y^{2} - (2 + λ) x + λ y = 0$
$∴$ Its center is $(\frac{2 + λ}{2}, \frac{- λ}{2})$
Since $A B$ is the diameter of the required circle therefore, the centre must lie on $A B$ ,
$\frac{2 + λ}{2} = \frac{- λ}{2}$
$\Rightarrow λ = - 1$
$∴$ The equation of the required circle is $x^{2} + y^{2} - 2 x - 1 (y - x) = 0$ $x^{2} + y^{2} - 2 x - y + x = 0$
$x^{2} + y^{2} - x - y = 0$

Hence

Conic Section

119637 Find the equation of a circle which touches both the axes and the line $3 x - 4 y + 8 = 0$ and lies in the third quadrant.

1

x^{2} + y^{2} + 4 x - 4 y - 4 = 0

2

x^{2} + y^{2} + 4 x + 4 y + 4 = 0

3

x^{2} + y^{2} - 4 x - 4 y + 4 = 0

4

x^{2} + y^{2} + 4 x + 4 y - 4 = 0

Explanation:

B Given circle is in third quadrant and touches both axes
Then centre of circle $(- a, - a)$ So, equation of circle,
original image
$(x + a)^{2} + (y + a)^{2} = a^{2}$
$x^{2} + y^{2} + 2 a x + 2 a y + a^{2} = 0$
Perpendicular distance from the centre of the circle
$| \frac{- 3 a + 4 a + 8}{5} | = a$
$| a + 8 | = 5 a$
$a + 8 = 5 a$
$a = 2$
$x^{2} + y^{2} + 2 a x + 2 a y + a^{2} = 0$
$Putting a = 2 in equation (i) -$
$x^{2} + y^{2} + 4 x + 4 y + 4 = 0$

COMEDK-2020

Conic Section

119633 If the radius of circle $x^{2} + y^{2} - 4 x + 6 y - k = 0$ is
5 , then $k =$

1 12

2 25

3 -12

4 -25

Explanation:

A : Given,
$x^{2} + y^{2} - 4 x + 6 y - k = 0$
$x^{2} + y^{2} - 4 x + 6 y + 4 - 4 + 9 - 9 - k = 0$
$(x^{2} - 4 x + 4) + (y^{2} + 6 y + 9) - (k + 4 + 9) = 0$
$(x - 2)^{2} + (y + 3)^{2} = (k + 13)$
$r^{2} = k + 13$
$5^{2} = k + 13$
$25 = k + 13$
$k = 12$

MHT CET-2020

Conic Section

119634 The radius of the circle passing through the points $(5, 7), (2, - 2)$ and $(- 2, 0)$ is

1 2 units

2 5 units

3 3 units

4 4 units

Explanation:

B original image
Given, Point $(5, 7), (2, - 2)$ and $(- 2, 0)$ lie on circle according to figure we can say that
Radius $OA = OC$
$[h - (- 2)]^{2} + (k)^{2} = (h - 5)^{2} + (k - 7)^{2}$
$h^{2} + 4 + 4 h + k^{2} = h^{2} + 25 - 10 h + k^{2} + 49 - 14 k$
$14 h + 14 k = 70$
$h + k = 5$
Similarly, Radius $OA = OB$
$(h + 2)^{2} + (k)^{2} = (h - 2)^{2} + (k + 2)^{2}$
$h^{2} + 4 + 4 h + k^{2} = h^{2} + 4 - 4 h + k^{2} + 4 + 4 k$
$4 h + 4 h - 4 k = 4$
$8 h - 4 k = 4$
$2 h - k = 1$
$2 h = 1 + k$
$By equation (i) and (ii)$
$We get h = 2, k = 3, centre of circle (2, 3)$
$Radius of circle = OA = \sqrt{(h + 2)^{2} + (k - 0)^{2}}$
$= \sqrt{(2 + 2)^{2} + 3^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5 units$

MHT CET-2020

Conic Section

119635 If the radius of the circle
$x^{2} + y^{2} - 18 x + 12 y + k = 0$ is 11 units, then the value of $k$ is

1 -3

2 4

3 3

4 -4

Explanation:

D Given, $r = 11$ units
We know that equation of circle is
$x^{2} + y^{2} + 2 gx + 2 fy + c = 0$
$Given, equation,$
$x^{2} + y^{2} - 18 x + 12 y + k = 0$
$Comparing with both equations.$
$2 g = - 18, 2 f = + 12, c = k$
$g = - 9, f = 6, c = k$
$We also know that$
$r = \sqrt{g^{2} + f^{2} - c}$
$11 = \sqrt{81 + 36 - k}, 11 = \sqrt{117 - k}$
$121 = 117 - k$
$k = - 4$

MHT CET-2019

Conic Section

119636 The equation of the circle concentric with the circle $x^{2} + y^{2} - 6 x - 4 y - 12 = 0$ and touching the $Y$ -axis is
#[Qdiff: Hard, QCat: Numerical Based, examname: Hence, $By$ comparing with standard equation\), $2 g = - 6, 2 f = - 4, c = - 12$ , $g = - 3, f = - 2, \(∴$ centre of the circle is $C (- g, - f) = (3, 2)$ . The required circle touches the $y -$ axis then radius of the, circle is 3., the equation of the circle is-, $(x - 3)^{2} + (y - 2)^{2} = 3^{2}$ , $x^{2} + 9 - 6 x + y^{2} + 4 - 4 y = 9$ , $x^{2} + y^{2} - 6 x - 4 y + 4 = 0$ , 14. The intercept on the line $y = x$ by the circle $x^{2} + y^{2} - 2 x = 0$ is $A B$ . The equation of the circle with $AB$ as a diameter is,

1

x^{2} + y^{2} - 6 x - 4 y - 4 = 0

2

x^{2} + y^{2} - 6 x - 4 y - 9 = 0

3

x^{2} + y^{2} - 6 x - 4 y + 9 = 0

4

x^{2} + y^{2} - 6 x - 4 y + 4 = 0

Explanation:

D Given,
$x^{2} + y^{2} - 6 x - 4 y - 12 = 0$
$A s w e k n o w t h a t e q u a t i o n o f a c i r c l e i s$
$x^{2} + y^{2} + 2 gx + 2 fy + c = 0$
Ans: a
Exp: (A) : In this question we will use the general equation of the circle passing through the points of intersection of any circle. $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ and any line $a x + b y + d = 0$ equation of a circle passing through the point of intersection of $x^{2} + y^{2} - 2 x = 0$ and $y = x$ is $x^{2} + y^{2} - 2 x + λ (y - x) = 0$ $x^{2} + y^{2} - (2 + λ) x + λ y = 0$
$∴$ Its center is $(\frac{2 + λ}{2}, \frac{- λ}{2})$
Since $A B$ is the diameter of the required circle therefore, the centre must lie on $A B$ ,
$\frac{2 + λ}{2} = \frac{- λ}{2}$
$\Rightarrow λ = - 1$
$∴$ The equation of the required circle is $x^{2} + y^{2} - 2 x - 1 (y - x) = 0$ $x^{2} + y^{2} - 2 x - y + x = 0$
$x^{2} + y^{2} - x - y = 0$

Hence

Conic Section

119637 Find the equation of a circle which touches both the axes and the line $3 x - 4 y + 8 = 0$ and lies in the third quadrant.

1

x^{2} + y^{2} + 4 x - 4 y - 4 = 0

2

x^{2} + y^{2} + 4 x + 4 y + 4 = 0

3

x^{2} + y^{2} - 4 x - 4 y + 4 = 0

4

x^{2} + y^{2} + 4 x + 4 y - 4 = 0

Explanation:

B Given circle is in third quadrant and touches both axes
Then centre of circle $(- a, - a)$ So, equation of circle,
original image
$(x + a)^{2} + (y + a)^{2} = a^{2}$
$x^{2} + y^{2} + 2 a x + 2 a y + a^{2} = 0$
Perpendicular distance from the centre of the circle
$| \frac{- 3 a + 4 a + 8}{5} | = a$
$| a + 8 | = 5 a$
$a + 8 = 5 a$
$a = 2$
$x^{2} + y^{2} + 2 a x + 2 a y + a^{2} = 0$
$Putting a = 2 in equation (i) -$
$x^{2} + y^{2} + 4 x + 4 y + 4 = 0$

COMEDK-2020

Conic Section

119633 If the radius of circle $x^{2} + y^{2} - 4 x + 6 y - k = 0$ is
5 , then $k =$

1 12

2 25

3 -12

4 -25

Explanation:

A : Given,
$x^{2} + y^{2} - 4 x + 6 y - k = 0$
$x^{2} + y^{2} - 4 x + 6 y + 4 - 4 + 9 - 9 - k = 0$
$(x^{2} - 4 x + 4) + (y^{2} + 6 y + 9) - (k + 4 + 9) = 0$
$(x - 2)^{2} + (y + 3)^{2} = (k + 13)$
$r^{2} = k + 13$
$5^{2} = k + 13$
$25 = k + 13$
$k = 12$

MHT CET-2020

Conic Section

119634 The radius of the circle passing through the points $(5, 7), (2, - 2)$ and $(- 2, 0)$ is

1 2 units

2 5 units

3 3 units

4 4 units

Explanation:

B original image
Given, Point $(5, 7), (2, - 2)$ and $(- 2, 0)$ lie on circle according to figure we can say that
Radius $OA = OC$
$[h - (- 2)]^{2} + (k)^{2} = (h - 5)^{2} + (k - 7)^{2}$
$h^{2} + 4 + 4 h + k^{2} = h^{2} + 25 - 10 h + k^{2} + 49 - 14 k$
$14 h + 14 k = 70$
$h + k = 5$
Similarly, Radius $OA = OB$
$(h + 2)^{2} + (k)^{2} = (h - 2)^{2} + (k + 2)^{2}$
$h^{2} + 4 + 4 h + k^{2} = h^{2} + 4 - 4 h + k^{2} + 4 + 4 k$
$4 h + 4 h - 4 k = 4$
$8 h - 4 k = 4$
$2 h - k = 1$
$2 h = 1 + k$
$By equation (i) and (ii)$
$We get h = 2, k = 3, centre of circle (2, 3)$
$Radius of circle = OA = \sqrt{(h + 2)^{2} + (k - 0)^{2}}$
$= \sqrt{(2 + 2)^{2} + 3^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5 units$

MHT CET-2020

Conic Section

119635 If the radius of the circle
$x^{2} + y^{2} - 18 x + 12 y + k = 0$ is 11 units, then the value of $k$ is

1 -3

2 4

3 3

4 -4

Explanation:

D Given, $r = 11$ units
We know that equation of circle is
$x^{2} + y^{2} + 2 gx + 2 fy + c = 0$
$Given, equation,$
$x^{2} + y^{2} - 18 x + 12 y + k = 0$
$Comparing with both equations.$
$2 g = - 18, 2 f = + 12, c = k$
$g = - 9, f = 6, c = k$
$We also know that$
$r = \sqrt{g^{2} + f^{2} - c}$
$11 = \sqrt{81 + 36 - k}, 11 = \sqrt{117 - k}$
$121 = 117 - k$
$k = - 4$

MHT CET-2019

Conic Section

119636 The equation of the circle concentric with the circle $x^{2} + y^{2} - 6 x - 4 y - 12 = 0$ and touching the $Y$ -axis is
#[Qdiff: Hard, QCat: Numerical Based, examname: Hence, $By$ comparing with standard equation\), $2 g = - 6, 2 f = - 4, c = - 12$ , $g = - 3, f = - 2, \(∴$ centre of the circle is $C (- g, - f) = (3, 2)$ . The required circle touches the $y -$ axis then radius of the, circle is 3., the equation of the circle is-, $(x - 3)^{2} + (y - 2)^{2} = 3^{2}$ , $x^{2} + 9 - 6 x + y^{2} + 4 - 4 y = 9$ , $x^{2} + y^{2} - 6 x - 4 y + 4 = 0$ , 14. The intercept on the line $y = x$ by the circle $x^{2} + y^{2} - 2 x = 0$ is $A B$ . The equation of the circle with $AB$ as a diameter is,

1

x^{2} + y^{2} - 6 x - 4 y - 4 = 0

2

x^{2} + y^{2} - 6 x - 4 y - 9 = 0

3

x^{2} + y^{2} - 6 x - 4 y + 9 = 0

4

x^{2} + y^{2} - 6 x - 4 y + 4 = 0

Explanation:

D Given,
$x^{2} + y^{2} - 6 x - 4 y - 12 = 0$
$A s w e k n o w t h a t e q u a t i o n o f a c i r c l e i s$
$x^{2} + y^{2} + 2 gx + 2 fy + c = 0$
Ans: a
Exp: (A) : In this question we will use the general equation of the circle passing through the points of intersection of any circle. $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ and any line $a x + b y + d = 0$ equation of a circle passing through the point of intersection of $x^{2} + y^{2} - 2 x = 0$ and $y = x$ is $x^{2} + y^{2} - 2 x + λ (y - x) = 0$ $x^{2} + y^{2} - (2 + λ) x + λ y = 0$
$∴$ Its center is $(\frac{2 + λ}{2}, \frac{- λ}{2})$
Since $A B$ is the diameter of the required circle therefore, the centre must lie on $A B$ ,
$\frac{2 + λ}{2} = \frac{- λ}{2}$
$\Rightarrow λ = - 1$
$∴$ The equation of the required circle is $x^{2} + y^{2} - 2 x - 1 (y - x) = 0$ $x^{2} + y^{2} - 2 x - y + x = 0$
$x^{2} + y^{2} - x - y = 0$

Hence

Conic Section

119637 Find the equation of a circle which touches both the axes and the line $3 x - 4 y + 8 = 0$ and lies in the third quadrant.

1

x^{2} + y^{2} + 4 x - 4 y - 4 = 0

2

x^{2} + y^{2} + 4 x + 4 y + 4 = 0

3

x^{2} + y^{2} - 4 x - 4 y + 4 = 0

4

x^{2} + y^{2} + 4 x + 4 y - 4 = 0

Explanation:

B Given circle is in third quadrant and touches both axes
Then centre of circle $(- a, - a)$ So, equation of circle,
original image
$(x + a)^{2} + (y + a)^{2} = a^{2}$
$x^{2} + y^{2} + 2 a x + 2 a y + a^{2} = 0$
Perpendicular distance from the centre of the circle
$| \frac{- 3 a + 4 a + 8}{5} | = a$
$| a + 8 | = 5 a$
$a + 8 = 5 a$
$a = 2$
$x^{2} + y^{2} + 2 a x + 2 a y + a^{2} = 0$
$Putting a = 2 in equation (i) -$
$x^{2} + y^{2} + 4 x + 4 y + 4 = 0$

COMEDK-2020