QBManager

Conic Section

120004 If $x^{2} + y^{2} + 6 x + 2 k y + 25 = 0$ to touch $Y$ -axis, then $k =$

1

\pm 20

2

- 1, - 5

3

\pm 5

4 4

Explanation:

C Given circle
$x^{2} + y^{2} + 6 x + 2 ky + 25 = 0$ touches $Y$ -axis
and here centre is $(- 3, - k)$ so radius $= 3$ units
Also radius $= \sqrt{9 + k^{2} - 25}$
$\Rightarrow \sqrt{k^{2} - 16} = 3 \Rightarrow k^{2} = 25 \Rightarrow k = \pm 5$

AP EAMCET-17.09.2020

Conic Section

120005 If one end of diameter of the circle $x^{2} + y^{2} - 4 x - 6 y + 11 = 0$ is $(3, 4)$ then the other end of the diameter is

1

(0, 1)

2

(1, 1)

3

(1, 2)

4

(1, 0)

Explanation:

C Given, circle $x^{2} + y^{2} - 4 x - 6 y + 11 = 0$
Center $= (2, 3)$
One end of diameter $= (3, 4)$
Let, other end be $(h, k)$
So, $\frac{h + 3}{2} = 2, \frac{k + 4}{2} = 3$
$h = 1, k = 2$
Hence, $(h, k) = (1, 2)$

AP EAMCET-17.09.2020

Conic Section

120014 The common tangent to the circles $x^{2} + y^{2} = 4$ and $x^{2} + y^{2} + 6 x + 8 y - 24 = 0$ also passes through the point

1

(6, - 2)

2

(4, - 2)

3

(- 6, 4)

4

(- 4, 6)

Explanation:

A Given two circle equation,
$S_{1} = x^{2} + y^{2} = 4$
With centre $C_{1} = (0, 0), r_{1} = 2$
$S_{2} = x^{2} + y^{2} + 6 x + 8 y - 24 = 0$
$(x + 3)^{2} + (y + 4)^{2} = 49$
$C_{2} = (- 3, - 4), r_{2} = 7$
$C_{1} C_{2} = | r_{1} - r_{2} |$
Equation of common tangent
$S_{1} - S_{2} = 0$
$6 x + 8 y - 20 = 0$
$3 x + 4 y = 10$ Hence, $(6, - 2)$ satisfies the equation.

JEE Main 09.04.2019

Conic Section

119978 The locus of the centre of a circle which touches externally the circle $x^{2} + y^{2} - 6 x - 6 y +$ $14 = 0$ and also touches the $y$ -axis is given by the equation

1

x^{2} - 6 x - 10 y + 14 = 0

2

x^{2} - 10 x - 6 y + 14 = 0

3

y^{2} - 6 x - 10 y + 14 = 0

4

y^{2} - 10 x - 6 y + 14 = 0

Explanation:

D Given,
original image
$x^{2} + y^{2} - 6 x - 6 y + 14 = 0$
$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
$by comparison of both equations$
$2 g = - 6, 2 f = - 6, C = 14$
$g = - 3,$
$(- g, - f) = (3, 3), f = - 3,$
$C = 14$
$r = \sqrt{(- 3)^{2} + (- 3)^{2} - 14}$
$= \sqrt{9 + 9 - 14} = \sqrt{4} = 2$
$∴ C_{1} C_{2} = h + 2$
$C_{1} C_{2} = \sqrt{(h - 3)^{2} + (k - 3)^{2}}$
$h + 2 = \sqrt{(h - 3)^{3} + (k - 3)^{2}}$
$h^{2} + 4 + 4 h = (h - 3)^{2} + (k - 3)^{2}$
$h^{2} + 4 + 4 h = h^{2} + 9 - 6 h + k^{2} + 9 - 6 k$
$k^{2} - 10 h - 6 k + 14 = 0$
$y^{2} - 10 x - 6 y + 14 = 0$
We know that,
From equation (i) and equation (ii), we get -
$h + 2 = \sqrt{(h - 3)^{3} + (k - 3)^{2}}$
$h^{2} + 4 + 4 h = (h - 3)^{2} + (k - 3)^{2}$
$h^{2} + 4 + 4 h = h^{2} + 9 - 6 h + k^{2} + 9 - 6 k$
$k^{2} - 10 h - 6 k + 14 = 0$
$y^{2} - 10 x - 6 y + 14 = 0$

JEEE-2018

Conic Section

119979 If the circles $x^{2} + y^{2} + 2 x + 2 k y + 6 = 0, x^{2} + y^{2} +$ $2 ky + k = 0$ intersect orthogonally, then $k$ : is

1 2 (or)

\frac{- 3}{2}

2 -2 (or)

\frac{- 3}{2}

3 2 (or)

\frac{3}{2}

4 -2 (or)

\frac{3}{2}

Explanation:

A Given,
$x^{2} + y^{2} + 2 x + 2 ky + 6 = 0$ $\dots .$ . (i)\)
$x^{2} + y^{2} + 2 ky + k = 0$ $\dots .$ . (ii)\)
We know that,
General equation of a circle,
$x^{2} + y^{2} + 2 gx + 2 fy + c = 0$
For equation (i), we get -
$2 g_{1} = 2,$ , $f = 2 k$ , $c_{1} = 6$
$g_{1} = 1, f_{1} = k$
For equation (ii), we get -
$2 g_{2} = 0, 2 f_{2} = 2 k$ , $c_{2} = k$
$g_{2} = 0 f_{2} = k$ $c_{2} = k$
Using condition of orthoganality
$2 f_{1} f_{2} + 2 g_{1} g_{2} = c_{1} + c_{2}$
$2 \times k \times k + 2 \times 1 \times 0 = 6 + k$
$2 k^{2} = 6 + k$
$2 k^{2} - k - 6 = 0$
$2 k^{2} - k - 6 = 0$
$2 k^{2} - 4 k + 3 k - 6 = 0$
$2 k (k - 2) + 3 (k - 2) = 0$
$k = 2, k = \frac{- 3}{2}$
$k = 2 or \frac{- 3}{2}$

JEEE-2019

Conic Section

120004 If $x^{2} + y^{2} + 6 x + 2 k y + 25 = 0$ to touch $Y$ -axis, then $k =$

1

\pm 20

2

- 1, - 5

3

\pm 5

4 4

Explanation:

C Given circle
$x^{2} + y^{2} + 6 x + 2 ky + 25 = 0$ touches $Y$ -axis
and here centre is $(- 3, - k)$ so radius $= 3$ units
Also radius $= \sqrt{9 + k^{2} - 25}$
$\Rightarrow \sqrt{k^{2} - 16} = 3 \Rightarrow k^{2} = 25 \Rightarrow k = \pm 5$

AP EAMCET-17.09.2020

Conic Section

120005 If one end of diameter of the circle $x^{2} + y^{2} - 4 x - 6 y + 11 = 0$ is $(3, 4)$ then the other end of the diameter is

1

(0, 1)

2

(1, 1)

3

(1, 2)

4

(1, 0)

Explanation:

C Given, circle $x^{2} + y^{2} - 4 x - 6 y + 11 = 0$
Center $= (2, 3)$
One end of diameter $= (3, 4)$
Let, other end be $(h, k)$
So, $\frac{h + 3}{2} = 2, \frac{k + 4}{2} = 3$
$h = 1, k = 2$
Hence, $(h, k) = (1, 2)$

AP EAMCET-17.09.2020

Conic Section

120014 The common tangent to the circles $x^{2} + y^{2} = 4$ and $x^{2} + y^{2} + 6 x + 8 y - 24 = 0$ also passes through the point

1

(6, - 2)

2

(4, - 2)

3

(- 6, 4)

4

(- 4, 6)

Explanation:

A Given two circle equation,
$S_{1} = x^{2} + y^{2} = 4$
With centre $C_{1} = (0, 0), r_{1} = 2$
$S_{2} = x^{2} + y^{2} + 6 x + 8 y - 24 = 0$
$(x + 3)^{2} + (y + 4)^{2} = 49$
$C_{2} = (- 3, - 4), r_{2} = 7$
$C_{1} C_{2} = | r_{1} - r_{2} |$
Equation of common tangent
$S_{1} - S_{2} = 0$
$6 x + 8 y - 20 = 0$
$3 x + 4 y = 10$ Hence, $(6, - 2)$ satisfies the equation.

JEE Main 09.04.2019

Conic Section

119978 The locus of the centre of a circle which touches externally the circle $x^{2} + y^{2} - 6 x - 6 y +$ $14 = 0$ and also touches the $y$ -axis is given by the equation

1

x^{2} - 6 x - 10 y + 14 = 0

2

x^{2} - 10 x - 6 y + 14 = 0

3

y^{2} - 6 x - 10 y + 14 = 0

4

y^{2} - 10 x - 6 y + 14 = 0

Explanation:

D Given,
original image
$x^{2} + y^{2} - 6 x - 6 y + 14 = 0$
$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
$by comparison of both equations$
$2 g = - 6, 2 f = - 6, C = 14$
$g = - 3,$
$(- g, - f) = (3, 3), f = - 3,$
$C = 14$
$r = \sqrt{(- 3)^{2} + (- 3)^{2} - 14}$
$= \sqrt{9 + 9 - 14} = \sqrt{4} = 2$
$∴ C_{1} C_{2} = h + 2$
$C_{1} C_{2} = \sqrt{(h - 3)^{2} + (k - 3)^{2}}$
$h + 2 = \sqrt{(h - 3)^{3} + (k - 3)^{2}}$
$h^{2} + 4 + 4 h = (h - 3)^{2} + (k - 3)^{2}$
$h^{2} + 4 + 4 h = h^{2} + 9 - 6 h + k^{2} + 9 - 6 k$
$k^{2} - 10 h - 6 k + 14 = 0$
$y^{2} - 10 x - 6 y + 14 = 0$
We know that,
From equation (i) and equation (ii), we get -
$h + 2 = \sqrt{(h - 3)^{3} + (k - 3)^{2}}$
$h^{2} + 4 + 4 h = (h - 3)^{2} + (k - 3)^{2}$
$h^{2} + 4 + 4 h = h^{2} + 9 - 6 h + k^{2} + 9 - 6 k$
$k^{2} - 10 h - 6 k + 14 = 0$
$y^{2} - 10 x - 6 y + 14 = 0$

JEEE-2018

Conic Section

119979 If the circles $x^{2} + y^{2} + 2 x + 2 k y + 6 = 0, x^{2} + y^{2} +$ $2 ky + k = 0$ intersect orthogonally, then $k$ : is

1 2 (or)

\frac{- 3}{2}

2 -2 (or)

\frac{- 3}{2}

3 2 (or)

\frac{3}{2}

4 -2 (or)

\frac{3}{2}

Explanation:

A Given,
$x^{2} + y^{2} + 2 x + 2 ky + 6 = 0$ $\dots .$ . (i)\)
$x^{2} + y^{2} + 2 ky + k = 0$ $\dots .$ . (ii)\)
We know that,
General equation of a circle,
$x^{2} + y^{2} + 2 gx + 2 fy + c = 0$
For equation (i), we get -
$2 g_{1} = 2,$ , $f = 2 k$ , $c_{1} = 6$
$g_{1} = 1, f_{1} = k$
For equation (ii), we get -
$2 g_{2} = 0, 2 f_{2} = 2 k$ , $c_{2} = k$
$g_{2} = 0 f_{2} = k$ $c_{2} = k$
Using condition of orthoganality
$2 f_{1} f_{2} + 2 g_{1} g_{2} = c_{1} + c_{2}$
$2 \times k \times k + 2 \times 1 \times 0 = 6 + k$
$2 k^{2} = 6 + k$
$2 k^{2} - k - 6 = 0$
$2 k^{2} - k - 6 = 0$
$2 k^{2} - 4 k + 3 k - 6 = 0$
$2 k (k - 2) + 3 (k - 2) = 0$
$k = 2, k = \frac{- 3}{2}$
$k = 2 or \frac{- 3}{2}$

JEEE-2019

Conic Section

120004 If $x^{2} + y^{2} + 6 x + 2 k y + 25 = 0$ to touch $Y$ -axis, then $k =$

1

\pm 20

2

- 1, - 5

3

\pm 5

4 4

Explanation:

C Given circle
$x^{2} + y^{2} + 6 x + 2 ky + 25 = 0$ touches $Y$ -axis
and here centre is $(- 3, - k)$ so radius $= 3$ units
Also radius $= \sqrt{9 + k^{2} - 25}$
$\Rightarrow \sqrt{k^{2} - 16} = 3 \Rightarrow k^{2} = 25 \Rightarrow k = \pm 5$

AP EAMCET-17.09.2020

Conic Section

120005 If one end of diameter of the circle $x^{2} + y^{2} - 4 x - 6 y + 11 = 0$ is $(3, 4)$ then the other end of the diameter is

1

(0, 1)

2

(1, 1)

3

(1, 2)

4

(1, 0)

Explanation:

C Given, circle $x^{2} + y^{2} - 4 x - 6 y + 11 = 0$
Center $= (2, 3)$
One end of diameter $= (3, 4)$
Let, other end be $(h, k)$
So, $\frac{h + 3}{2} = 2, \frac{k + 4}{2} = 3$
$h = 1, k = 2$
Hence, $(h, k) = (1, 2)$

AP EAMCET-17.09.2020

Conic Section

120014 The common tangent to the circles $x^{2} + y^{2} = 4$ and $x^{2} + y^{2} + 6 x + 8 y - 24 = 0$ also passes through the point

1

(6, - 2)

2

(4, - 2)

3

(- 6, 4)

4

(- 4, 6)

Explanation:

A Given two circle equation,
$S_{1} = x^{2} + y^{2} = 4$
With centre $C_{1} = (0, 0), r_{1} = 2$
$S_{2} = x^{2} + y^{2} + 6 x + 8 y - 24 = 0$
$(x + 3)^{2} + (y + 4)^{2} = 49$
$C_{2} = (- 3, - 4), r_{2} = 7$
$C_{1} C_{2} = | r_{1} - r_{2} |$
Equation of common tangent
$S_{1} - S_{2} = 0$
$6 x + 8 y - 20 = 0$
$3 x + 4 y = 10$ Hence, $(6, - 2)$ satisfies the equation.

JEE Main 09.04.2019

Conic Section

119978 The locus of the centre of a circle which touches externally the circle $x^{2} + y^{2} - 6 x - 6 y +$ $14 = 0$ and also touches the $y$ -axis is given by the equation

1

x^{2} - 6 x - 10 y + 14 = 0

2

x^{2} - 10 x - 6 y + 14 = 0

3

y^{2} - 6 x - 10 y + 14 = 0

4

y^{2} - 10 x - 6 y + 14 = 0

Explanation:

D Given,
original image
$x^{2} + y^{2} - 6 x - 6 y + 14 = 0$
$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
$by comparison of both equations$
$2 g = - 6, 2 f = - 6, C = 14$
$g = - 3,$
$(- g, - f) = (3, 3), f = - 3,$
$C = 14$
$r = \sqrt{(- 3)^{2} + (- 3)^{2} - 14}$
$= \sqrt{9 + 9 - 14} = \sqrt{4} = 2$
$∴ C_{1} C_{2} = h + 2$
$C_{1} C_{2} = \sqrt{(h - 3)^{2} + (k - 3)^{2}}$
$h + 2 = \sqrt{(h - 3)^{3} + (k - 3)^{2}}$
$h^{2} + 4 + 4 h = (h - 3)^{2} + (k - 3)^{2}$
$h^{2} + 4 + 4 h = h^{2} + 9 - 6 h + k^{2} + 9 - 6 k$
$k^{2} - 10 h - 6 k + 14 = 0$
$y^{2} - 10 x - 6 y + 14 = 0$
We know that,
From equation (i) and equation (ii), we get -
$h + 2 = \sqrt{(h - 3)^{3} + (k - 3)^{2}}$
$h^{2} + 4 + 4 h = (h - 3)^{2} + (k - 3)^{2}$
$h^{2} + 4 + 4 h = h^{2} + 9 - 6 h + k^{2} + 9 - 6 k$
$k^{2} - 10 h - 6 k + 14 = 0$
$y^{2} - 10 x - 6 y + 14 = 0$

JEEE-2018

Conic Section

119979 If the circles $x^{2} + y^{2} + 2 x + 2 k y + 6 = 0, x^{2} + y^{2} +$ $2 ky + k = 0$ intersect orthogonally, then $k$ : is

1 2 (or)

\frac{- 3}{2}

2 -2 (or)

\frac{- 3}{2}

3 2 (or)

\frac{3}{2}

4 -2 (or)

\frac{3}{2}

Explanation:

A Given,
$x^{2} + y^{2} + 2 x + 2 ky + 6 = 0$ $\dots .$ . (i)\)
$x^{2} + y^{2} + 2 ky + k = 0$ $\dots .$ . (ii)\)
We know that,
General equation of a circle,
$x^{2} + y^{2} + 2 gx + 2 fy + c = 0$
For equation (i), we get -
$2 g_{1} = 2,$ , $f = 2 k$ , $c_{1} = 6$
$g_{1} = 1, f_{1} = k$
For equation (ii), we get -
$2 g_{2} = 0, 2 f_{2} = 2 k$ , $c_{2} = k$
$g_{2} = 0 f_{2} = k$ $c_{2} = k$
Using condition of orthoganality
$2 f_{1} f_{2} + 2 g_{1} g_{2} = c_{1} + c_{2}$
$2 \times k \times k + 2 \times 1 \times 0 = 6 + k$
$2 k^{2} = 6 + k$
$2 k^{2} - k - 6 = 0$
$2 k^{2} - k - 6 = 0$
$2 k^{2} - 4 k + 3 k - 6 = 0$
$2 k (k - 2) + 3 (k - 2) = 0$
$k = 2, k = \frac{- 3}{2}$
$k = 2 or \frac{- 3}{2}$

JEEE-2019

Conic Section

120004 If $x^{2} + y^{2} + 6 x + 2 k y + 25 = 0$ to touch $Y$ -axis, then $k =$

1

\pm 20

2

- 1, - 5

3

\pm 5

4 4

Explanation:

C Given circle
$x^{2} + y^{2} + 6 x + 2 ky + 25 = 0$ touches $Y$ -axis
and here centre is $(- 3, - k)$ so radius $= 3$ units
Also radius $= \sqrt{9 + k^{2} - 25}$
$\Rightarrow \sqrt{k^{2} - 16} = 3 \Rightarrow k^{2} = 25 \Rightarrow k = \pm 5$

AP EAMCET-17.09.2020

Conic Section

120005 If one end of diameter of the circle $x^{2} + y^{2} - 4 x - 6 y + 11 = 0$ is $(3, 4)$ then the other end of the diameter is

1

(0, 1)

2

(1, 1)

3

(1, 2)

4

(1, 0)

Explanation:

C Given, circle $x^{2} + y^{2} - 4 x - 6 y + 11 = 0$
Center $= (2, 3)$
One end of diameter $= (3, 4)$
Let, other end be $(h, k)$
So, $\frac{h + 3}{2} = 2, \frac{k + 4}{2} = 3$
$h = 1, k = 2$
Hence, $(h, k) = (1, 2)$

AP EAMCET-17.09.2020

Conic Section

120014 The common tangent to the circles $x^{2} + y^{2} = 4$ and $x^{2} + y^{2} + 6 x + 8 y - 24 = 0$ also passes through the point

1

(6, - 2)

2

(4, - 2)

3

(- 6, 4)

4

(- 4, 6)

Explanation:

A Given two circle equation,
$S_{1} = x^{2} + y^{2} = 4$
With centre $C_{1} = (0, 0), r_{1} = 2$
$S_{2} = x^{2} + y^{2} + 6 x + 8 y - 24 = 0$
$(x + 3)^{2} + (y + 4)^{2} = 49$
$C_{2} = (- 3, - 4), r_{2} = 7$
$C_{1} C_{2} = | r_{1} - r_{2} |$
Equation of common tangent
$S_{1} - S_{2} = 0$
$6 x + 8 y - 20 = 0$
$3 x + 4 y = 10$ Hence, $(6, - 2)$ satisfies the equation.

JEE Main 09.04.2019

Conic Section

119978 The locus of the centre of a circle which touches externally the circle $x^{2} + y^{2} - 6 x - 6 y +$ $14 = 0$ and also touches the $y$ -axis is given by the equation

1

x^{2} - 6 x - 10 y + 14 = 0

2

x^{2} - 10 x - 6 y + 14 = 0

3

y^{2} - 6 x - 10 y + 14 = 0

4

y^{2} - 10 x - 6 y + 14 = 0

Explanation:

D Given,
original image
$x^{2} + y^{2} - 6 x - 6 y + 14 = 0$
$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
$by comparison of both equations$
$2 g = - 6, 2 f = - 6, C = 14$
$g = - 3,$
$(- g, - f) = (3, 3), f = - 3,$
$C = 14$
$r = \sqrt{(- 3)^{2} + (- 3)^{2} - 14}$
$= \sqrt{9 + 9 - 14} = \sqrt{4} = 2$
$∴ C_{1} C_{2} = h + 2$
$C_{1} C_{2} = \sqrt{(h - 3)^{2} + (k - 3)^{2}}$
$h + 2 = \sqrt{(h - 3)^{3} + (k - 3)^{2}}$
$h^{2} + 4 + 4 h = (h - 3)^{2} + (k - 3)^{2}$
$h^{2} + 4 + 4 h = h^{2} + 9 - 6 h + k^{2} + 9 - 6 k$
$k^{2} - 10 h - 6 k + 14 = 0$
$y^{2} - 10 x - 6 y + 14 = 0$
We know that,
From equation (i) and equation (ii), we get -
$h + 2 = \sqrt{(h - 3)^{3} + (k - 3)^{2}}$
$h^{2} + 4 + 4 h = (h - 3)^{2} + (k - 3)^{2}$
$h^{2} + 4 + 4 h = h^{2} + 9 - 6 h + k^{2} + 9 - 6 k$
$k^{2} - 10 h - 6 k + 14 = 0$
$y^{2} - 10 x - 6 y + 14 = 0$

JEEE-2018

Conic Section

119979 If the circles $x^{2} + y^{2} + 2 x + 2 k y + 6 = 0, x^{2} + y^{2} +$ $2 ky + k = 0$ intersect orthogonally, then $k$ : is

1 2 (or)

\frac{- 3}{2}

2 -2 (or)

\frac{- 3}{2}

3 2 (or)

\frac{3}{2}

4 -2 (or)

\frac{3}{2}

Explanation:

A Given,
$x^{2} + y^{2} + 2 x + 2 ky + 6 = 0$ $\dots .$ . (i)\)
$x^{2} + y^{2} + 2 ky + k = 0$ $\dots .$ . (ii)\)
We know that,
General equation of a circle,
$x^{2} + y^{2} + 2 gx + 2 fy + c = 0$
For equation (i), we get -
$2 g_{1} = 2,$ , $f = 2 k$ , $c_{1} = 6$
$g_{1} = 1, f_{1} = k$
For equation (ii), we get -
$2 g_{2} = 0, 2 f_{2} = 2 k$ , $c_{2} = k$
$g_{2} = 0 f_{2} = k$ $c_{2} = k$
Using condition of orthoganality
$2 f_{1} f_{2} + 2 g_{1} g_{2} = c_{1} + c_{2}$
$2 \times k \times k + 2 \times 1 \times 0 = 6 + k$
$2 k^{2} = 6 + k$
$2 k^{2} - k - 6 = 0$
$2 k^{2} - k - 6 = 0$
$2 k^{2} - 4 k + 3 k - 6 = 0$
$2 k (k - 2) + 3 (k - 2) = 0$
$k = 2, k = \frac{- 3}{2}$
$k = 2 or \frac{- 3}{2}$

JEEE-2019

Conic Section

120004 If $x^{2} + y^{2} + 6 x + 2 k y + 25 = 0$ to touch $Y$ -axis, then $k =$

1

\pm 20

2

- 1, - 5

3

\pm 5

4 4

Explanation:

C Given circle
$x^{2} + y^{2} + 6 x + 2 ky + 25 = 0$ touches $Y$ -axis
and here centre is $(- 3, - k)$ so radius $= 3$ units
Also radius $= \sqrt{9 + k^{2} - 25}$
$\Rightarrow \sqrt{k^{2} - 16} = 3 \Rightarrow k^{2} = 25 \Rightarrow k = \pm 5$

AP EAMCET-17.09.2020

Conic Section

120005 If one end of diameter of the circle $x^{2} + y^{2} - 4 x - 6 y + 11 = 0$ is $(3, 4)$ then the other end of the diameter is

1

(0, 1)

2

(1, 1)

3

(1, 2)

4

(1, 0)

Explanation:

C Given, circle $x^{2} + y^{2} - 4 x - 6 y + 11 = 0$
Center $= (2, 3)$
One end of diameter $= (3, 4)$
Let, other end be $(h, k)$
So, $\frac{h + 3}{2} = 2, \frac{k + 4}{2} = 3$
$h = 1, k = 2$
Hence, $(h, k) = (1, 2)$

AP EAMCET-17.09.2020

Conic Section

120014 The common tangent to the circles $x^{2} + y^{2} = 4$ and $x^{2} + y^{2} + 6 x + 8 y - 24 = 0$ also passes through the point

1

(6, - 2)

2

(4, - 2)

3

(- 6, 4)

4

(- 4, 6)

Explanation:

A Given two circle equation,
$S_{1} = x^{2} + y^{2} = 4$
With centre $C_{1} = (0, 0), r_{1} = 2$
$S_{2} = x^{2} + y^{2} + 6 x + 8 y - 24 = 0$
$(x + 3)^{2} + (y + 4)^{2} = 49$
$C_{2} = (- 3, - 4), r_{2} = 7$
$C_{1} C_{2} = | r_{1} - r_{2} |$
Equation of common tangent
$S_{1} - S_{2} = 0$
$6 x + 8 y - 20 = 0$
$3 x + 4 y = 10$ Hence, $(6, - 2)$ satisfies the equation.

JEE Main 09.04.2019

Conic Section

119978 The locus of the centre of a circle which touches externally the circle $x^{2} + y^{2} - 6 x - 6 y +$ $14 = 0$ and also touches the $y$ -axis is given by the equation

1

x^{2} - 6 x - 10 y + 14 = 0

2

x^{2} - 10 x - 6 y + 14 = 0

3

y^{2} - 6 x - 10 y + 14 = 0

4

y^{2} - 10 x - 6 y + 14 = 0

Explanation:

D Given,
original image
$x^{2} + y^{2} - 6 x - 6 y + 14 = 0$
$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
$by comparison of both equations$
$2 g = - 6, 2 f = - 6, C = 14$
$g = - 3,$
$(- g, - f) = (3, 3), f = - 3,$
$C = 14$
$r = \sqrt{(- 3)^{2} + (- 3)^{2} - 14}$
$= \sqrt{9 + 9 - 14} = \sqrt{4} = 2$
$∴ C_{1} C_{2} = h + 2$
$C_{1} C_{2} = \sqrt{(h - 3)^{2} + (k - 3)^{2}}$
$h + 2 = \sqrt{(h - 3)^{3} + (k - 3)^{2}}$
$h^{2} + 4 + 4 h = (h - 3)^{2} + (k - 3)^{2}$
$h^{2} + 4 + 4 h = h^{2} + 9 - 6 h + k^{2} + 9 - 6 k$
$k^{2} - 10 h - 6 k + 14 = 0$
$y^{2} - 10 x - 6 y + 14 = 0$
We know that,
From equation (i) and equation (ii), we get -
$h + 2 = \sqrt{(h - 3)^{3} + (k - 3)^{2}}$
$h^{2} + 4 + 4 h = (h - 3)^{2} + (k - 3)^{2}$
$h^{2} + 4 + 4 h = h^{2} + 9 - 6 h + k^{2} + 9 - 6 k$
$k^{2} - 10 h - 6 k + 14 = 0$
$y^{2} - 10 x - 6 y + 14 = 0$

JEEE-2018

Conic Section

119979 If the circles $x^{2} + y^{2} + 2 x + 2 k y + 6 = 0, x^{2} + y^{2} +$ $2 ky + k = 0$ intersect orthogonally, then $k$ : is

1 2 (or)

\frac{- 3}{2}

2 -2 (or)

\frac{- 3}{2}

3 2 (or)

\frac{3}{2}

4 -2 (or)

\frac{3}{2}

Explanation:

A Given,
$x^{2} + y^{2} + 2 x + 2 ky + 6 = 0$ $\dots .$ . (i)\)
$x^{2} + y^{2} + 2 ky + k = 0$ $\dots .$ . (ii)\)
We know that,
General equation of a circle,
$x^{2} + y^{2} + 2 gx + 2 fy + c = 0$
For equation (i), we get -
$2 g_{1} = 2,$ , $f = 2 k$ , $c_{1} = 6$
$g_{1} = 1, f_{1} = k$
For equation (ii), we get -
$2 g_{2} = 0, 2 f_{2} = 2 k$ , $c_{2} = k$
$g_{2} = 0 f_{2} = k$ $c_{2} = k$
Using condition of orthoganality
$2 f_{1} f_{2} + 2 g_{1} g_{2} = c_{1} + c_{2}$
$2 \times k \times k + 2 \times 1 \times 0 = 6 + k$
$2 k^{2} = 6 + k$
$2 k^{2} - k - 6 = 0$
$2 k^{2} - k - 6 = 0$
$2 k^{2} - 4 k + 3 k - 6 = 0$
$2 k (k - 2) + 3 (k - 2) = 0$
$k = 2, k = \frac{- 3}{2}$
$k = 2 or \frac{- 3}{2}$

JEEE-2019

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