QBManager

Conic Section

119861 The number of common tangents to the circles $x^{2} + y^{2} = 4$ and $x^{2} + y^{2} - 6 x - 8 y - 24 = 0$ is

1 3

2 4

3 2

4 1

Explanation:

D $: x^{2} + y^{2} = 4$
$Centre c_{1} (0, 0), r_{1} = 2$
$x^{2} + y^{2} - 6 x - 8 y - 24 = 0$
$2 g = - 6, 2 f = - 8, c = - 24$
$g = - 3, f = - 4, c = - 24$
$(- g, - f) = (3, 4) r_{2} = \sqrt{(+ 3)^{2} + (4)^{2} - (- 24)}$
$= \sqrt{9 + 16 + 24} = \sqrt{49} = 7$
$c_{2} = (3, 4), r_{2} = 7$
$c_{1} c_{2} = \sqrt{(0 - 3)^{2} + (0 - 4)^{2}}$
$= \sqrt{9 + 16} = \sqrt{25} = 5$
$∴ c_{1} c_{2} = r_{2} - r_{1}$
$5 = 7 - 2 = 5$
$5 = 5$
Hence the given circles touch each other internally. Hence the number of common tangent is only one

BCECE-2012

Conic Section

119862 The circle $S_{1}$ with centre $C_{1} (a_{1}, b_{1})$ and radius $r_{1}$ touches externally the circle $S_{2}$ with centre $C_{2} (a_{2}, b_{2})$ and radius $r_{2}$ . If the tangent at their common point passes through the origin, then

1

(a_{1}^{2} + a_{2}^{2}) + (b_{1}^{2} + b_{2}^{2}) = r_{1}^{2} + r_{2}^{2}

2

(a_{1}^{2} - a_{2}^{2}) + (b_{1}^{2} - b_{2}^{2}) = r_{1}^{2} - r_{2}^{2}

3

{(a_{1}^{2} - b_{2})}^{2} + (a_{2}^{2} + b_{2}^{2}) = r_{1}^{2} + r_{2}^{2}

4

(a_{1}^{2} - b_{1}^{2}) + (a_{1}^{2} + b_{2}^{2}) = r_{1}^{2} + r_{2}^{2}

Explanation:

B According to question we have
$S_{1} = {(x - a_{1})}^{2} + {(y - b_{1})}^{2} = r_{1}^{2}$
$S_{2} = {(x - a_{2})}^{2} + {(y - b_{2})}^{2} = r_{2}^{2}$
$for common tangent,$
$S_{1} - S_{2} = 0$
$2 x (a_{1} - a_{2}) + 2 y (b_{1} - b_{2}) + (a_{2}^{2} + b_{2}^{2}) - (a_{1}^{2} + b_{1}^{2}) + r_{1}^{2}$
$- r_{2}^{2} = 0$
$if this passes through the origin then$
$(a_{2}^{2} + b_{2}^{2}) - (a_{1}^{2} + b_{1}^{2}) + r_{1}^{2} - r_{2}^{2} = 0$
$(a_{2}^{2} - a_{1}^{2}) + (b_{2}^{2} - b_{1}^{2}) = r_{2}^{2} - r_{1}^{2}$
${(a_{1}^{2} - a_{2})}^{2}) + (b_{1}^{2} - b_{2}^{2}) = (r_{1}^{2} - r_{2}^{2})$

BCECE-2011

Conic Section

119863 The equation of a circle of radius 5 which lies within the circle $x^{2} + y^{2} + 14 x + 10 y - 26 = 0$ and touches it at the point $(- 1, 3)$ is

1

x^{2} + y^{2} + 8 x + 2 y + 8 = 0

2

x^{2} + y^{2} + 8 x + 2 y - 8 = 0

3

x^{2} + y^{2} + 8 x + 2 y - 14 = 0

4 None of the above

Explanation:

B $x^{2} + y^{2} + 14 x + 10 y - 26 = 0$
$2 g = 14, 2 f = 10, c = - 26$
$g = 7, f = + 5, c = - 26$
$(- g, - f) = (- 7, - 5), r = \sqrt{(7)^{2} + (5)^{2} - (- 26)}$
$= \sqrt{49 + 25 + 26}$
$= \sqrt{100} = 10$
$(- 4, - 10)$
$r = 5$
According to question, other circle lies in the circle touching it at $(- 1, 3)$ and its radius is 5 , So it also passes through the centre of the given circle which is $(- 7, - 5)$ Now we have the ends point of diameter of the circle
So, its centre is $(\frac{- 7 - 1}{2}, \frac{- 5 + 3}{2})$
$r = 5$
So the equation of the other circle is $(x + 4)^{2} + (y + 1)^{2} = 5^{2}$ $x^{2} + 16 + 8 x + y^{2} + 1 + 2 y = 25$ $x^{2} + y^{2} + 8 x + 2 y - 8 = 0$

BCECE-2010

Conic Section

119864 The number of common tangents to two circles $x^{2} + y^{2} = 4$ and $x^{2} + y^{2} - 8 x + 12 = 0$ is

1 1

2 2

3 3

4 4

Explanation:

C Given equations of circles are
$x^{2} + y^{2} = 4$
and $x^{2} + y^{2} - 8 x + 12 = 0$
or
$x^{2} + y^{2} = (2)^{2}$
$(x - 4)^{2} + (y - 0)^{2} = (2)^{2}$
The figure of both the circles are shown blow:
original image
From figure, we see that there is exactly three common tangents.

CG PET- 2015

Conic Section

119861 The number of common tangents to the circles $x^{2} + y^{2} = 4$ and $x^{2} + y^{2} - 6 x - 8 y - 24 = 0$ is

1 3

2 4

3 2

4 1

Explanation:

D $: x^{2} + y^{2} = 4$
$Centre c_{1} (0, 0), r_{1} = 2$
$x^{2} + y^{2} - 6 x - 8 y - 24 = 0$
$2 g = - 6, 2 f = - 8, c = - 24$
$g = - 3, f = - 4, c = - 24$
$(- g, - f) = (3, 4) r_{2} = \sqrt{(+ 3)^{2} + (4)^{2} - (- 24)}$
$= \sqrt{9 + 16 + 24} = \sqrt{49} = 7$
$c_{2} = (3, 4), r_{2} = 7$
$c_{1} c_{2} = \sqrt{(0 - 3)^{2} + (0 - 4)^{2}}$
$= \sqrt{9 + 16} = \sqrt{25} = 5$
$∴ c_{1} c_{2} = r_{2} - r_{1}$
$5 = 7 - 2 = 5$
$5 = 5$
Hence the given circles touch each other internally. Hence the number of common tangent is only one

BCECE-2012

Conic Section

119862 The circle $S_{1}$ with centre $C_{1} (a_{1}, b_{1})$ and radius $r_{1}$ touches externally the circle $S_{2}$ with centre $C_{2} (a_{2}, b_{2})$ and radius $r_{2}$ . If the tangent at their common point passes through the origin, then

1

(a_{1}^{2} + a_{2}^{2}) + (b_{1}^{2} + b_{2}^{2}) = r_{1}^{2} + r_{2}^{2}

2

(a_{1}^{2} - a_{2}^{2}) + (b_{1}^{2} - b_{2}^{2}) = r_{1}^{2} - r_{2}^{2}

3

{(a_{1}^{2} - b_{2})}^{2} + (a_{2}^{2} + b_{2}^{2}) = r_{1}^{2} + r_{2}^{2}

4

(a_{1}^{2} - b_{1}^{2}) + (a_{1}^{2} + b_{2}^{2}) = r_{1}^{2} + r_{2}^{2}

Explanation:

B According to question we have
$S_{1} = {(x - a_{1})}^{2} + {(y - b_{1})}^{2} = r_{1}^{2}$
$S_{2} = {(x - a_{2})}^{2} + {(y - b_{2})}^{2} = r_{2}^{2}$
$for common tangent,$
$S_{1} - S_{2} = 0$
$2 x (a_{1} - a_{2}) + 2 y (b_{1} - b_{2}) + (a_{2}^{2} + b_{2}^{2}) - (a_{1}^{2} + b_{1}^{2}) + r_{1}^{2}$
$- r_{2}^{2} = 0$
$if this passes through the origin then$
$(a_{2}^{2} + b_{2}^{2}) - (a_{1}^{2} + b_{1}^{2}) + r_{1}^{2} - r_{2}^{2} = 0$
$(a_{2}^{2} - a_{1}^{2}) + (b_{2}^{2} - b_{1}^{2}) = r_{2}^{2} - r_{1}^{2}$
${(a_{1}^{2} - a_{2})}^{2}) + (b_{1}^{2} - b_{2}^{2}) = (r_{1}^{2} - r_{2}^{2})$

BCECE-2011

Conic Section

119863 The equation of a circle of radius 5 which lies within the circle $x^{2} + y^{2} + 14 x + 10 y - 26 = 0$ and touches it at the point $(- 1, 3)$ is

1

x^{2} + y^{2} + 8 x + 2 y + 8 = 0

2

x^{2} + y^{2} + 8 x + 2 y - 8 = 0

3

x^{2} + y^{2} + 8 x + 2 y - 14 = 0

4 None of the above

Explanation:

B $x^{2} + y^{2} + 14 x + 10 y - 26 = 0$
$2 g = 14, 2 f = 10, c = - 26$
$g = 7, f = + 5, c = - 26$
$(- g, - f) = (- 7, - 5), r = \sqrt{(7)^{2} + (5)^{2} - (- 26)}$
$= \sqrt{49 + 25 + 26}$
$= \sqrt{100} = 10$
$(- 4, - 10)$
$r = 5$
According to question, other circle lies in the circle touching it at $(- 1, 3)$ and its radius is 5 , So it also passes through the centre of the given circle which is $(- 7, - 5)$ Now we have the ends point of diameter of the circle
So, its centre is $(\frac{- 7 - 1}{2}, \frac{- 5 + 3}{2})$
$r = 5$
So the equation of the other circle is $(x + 4)^{2} + (y + 1)^{2} = 5^{2}$ $x^{2} + 16 + 8 x + y^{2} + 1 + 2 y = 25$ $x^{2} + y^{2} + 8 x + 2 y - 8 = 0$

BCECE-2010

Conic Section

119864 The number of common tangents to two circles $x^{2} + y^{2} = 4$ and $x^{2} + y^{2} - 8 x + 12 = 0$ is

1 1

2 2

3 3

4 4

Explanation:

C Given equations of circles are
$x^{2} + y^{2} = 4$
and $x^{2} + y^{2} - 8 x + 12 = 0$
or
$x^{2} + y^{2} = (2)^{2}$
$(x - 4)^{2} + (y - 0)^{2} = (2)^{2}$
The figure of both the circles are shown blow:
original image
From figure, we see that there is exactly three common tangents.

CG PET- 2015

Conic Section

119861 The number of common tangents to the circles $x^{2} + y^{2} = 4$ and $x^{2} + y^{2} - 6 x - 8 y - 24 = 0$ is

1 3

2 4

3 2

4 1

Explanation:

D $: x^{2} + y^{2} = 4$
$Centre c_{1} (0, 0), r_{1} = 2$
$x^{2} + y^{2} - 6 x - 8 y - 24 = 0$
$2 g = - 6, 2 f = - 8, c = - 24$
$g = - 3, f = - 4, c = - 24$
$(- g, - f) = (3, 4) r_{2} = \sqrt{(+ 3)^{2} + (4)^{2} - (- 24)}$
$= \sqrt{9 + 16 + 24} = \sqrt{49} = 7$
$c_{2} = (3, 4), r_{2} = 7$
$c_{1} c_{2} = \sqrt{(0 - 3)^{2} + (0 - 4)^{2}}$
$= \sqrt{9 + 16} = \sqrt{25} = 5$
$∴ c_{1} c_{2} = r_{2} - r_{1}$
$5 = 7 - 2 = 5$
$5 = 5$
Hence the given circles touch each other internally. Hence the number of common tangent is only one

BCECE-2012

Conic Section

119862 The circle $S_{1}$ with centre $C_{1} (a_{1}, b_{1})$ and radius $r_{1}$ touches externally the circle $S_{2}$ with centre $C_{2} (a_{2}, b_{2})$ and radius $r_{2}$ . If the tangent at their common point passes through the origin, then

1

(a_{1}^{2} + a_{2}^{2}) + (b_{1}^{2} + b_{2}^{2}) = r_{1}^{2} + r_{2}^{2}

2

(a_{1}^{2} - a_{2}^{2}) + (b_{1}^{2} - b_{2}^{2}) = r_{1}^{2} - r_{2}^{2}

3

{(a_{1}^{2} - b_{2})}^{2} + (a_{2}^{2} + b_{2}^{2}) = r_{1}^{2} + r_{2}^{2}

4

(a_{1}^{2} - b_{1}^{2}) + (a_{1}^{2} + b_{2}^{2}) = r_{1}^{2} + r_{2}^{2}

Explanation:

B According to question we have
$S_{1} = {(x - a_{1})}^{2} + {(y - b_{1})}^{2} = r_{1}^{2}$
$S_{2} = {(x - a_{2})}^{2} + {(y - b_{2})}^{2} = r_{2}^{2}$
$for common tangent,$
$S_{1} - S_{2} = 0$
$2 x (a_{1} - a_{2}) + 2 y (b_{1} - b_{2}) + (a_{2}^{2} + b_{2}^{2}) - (a_{1}^{2} + b_{1}^{2}) + r_{1}^{2}$
$- r_{2}^{2} = 0$
$if this passes through the origin then$
$(a_{2}^{2} + b_{2}^{2}) - (a_{1}^{2} + b_{1}^{2}) + r_{1}^{2} - r_{2}^{2} = 0$
$(a_{2}^{2} - a_{1}^{2}) + (b_{2}^{2} - b_{1}^{2}) = r_{2}^{2} - r_{1}^{2}$
${(a_{1}^{2} - a_{2})}^{2}) + (b_{1}^{2} - b_{2}^{2}) = (r_{1}^{2} - r_{2}^{2})$

BCECE-2011

Conic Section

119863 The equation of a circle of radius 5 which lies within the circle $x^{2} + y^{2} + 14 x + 10 y - 26 = 0$ and touches it at the point $(- 1, 3)$ is

1

x^{2} + y^{2} + 8 x + 2 y + 8 = 0

2

x^{2} + y^{2} + 8 x + 2 y - 8 = 0

3

x^{2} + y^{2} + 8 x + 2 y - 14 = 0

4 None of the above

Explanation:

B $x^{2} + y^{2} + 14 x + 10 y - 26 = 0$
$2 g = 14, 2 f = 10, c = - 26$
$g = 7, f = + 5, c = - 26$
$(- g, - f) = (- 7, - 5), r = \sqrt{(7)^{2} + (5)^{2} - (- 26)}$
$= \sqrt{49 + 25 + 26}$
$= \sqrt{100} = 10$
$(- 4, - 10)$
$r = 5$
According to question, other circle lies in the circle touching it at $(- 1, 3)$ and its radius is 5 , So it also passes through the centre of the given circle which is $(- 7, - 5)$ Now we have the ends point of diameter of the circle
So, its centre is $(\frac{- 7 - 1}{2}, \frac{- 5 + 3}{2})$
$r = 5$
So the equation of the other circle is $(x + 4)^{2} + (y + 1)^{2} = 5^{2}$ $x^{2} + 16 + 8 x + y^{2} + 1 + 2 y = 25$ $x^{2} + y^{2} + 8 x + 2 y - 8 = 0$

BCECE-2010

Conic Section

119864 The number of common tangents to two circles $x^{2} + y^{2} = 4$ and $x^{2} + y^{2} - 8 x + 12 = 0$ is

1 1

2 2

3 3

4 4

Explanation:

C Given equations of circles are
$x^{2} + y^{2} = 4$
and $x^{2} + y^{2} - 8 x + 12 = 0$
or
$x^{2} + y^{2} = (2)^{2}$
$(x - 4)^{2} + (y - 0)^{2} = (2)^{2}$
The figure of both the circles are shown blow:
original image
From figure, we see that there is exactly three common tangents.

CG PET- 2015

Conic Section

119861 The number of common tangents to the circles $x^{2} + y^{2} = 4$ and $x^{2} + y^{2} - 6 x - 8 y - 24 = 0$ is

1 3

2 4

3 2

4 1

Explanation:

D $: x^{2} + y^{2} = 4$
$Centre c_{1} (0, 0), r_{1} = 2$
$x^{2} + y^{2} - 6 x - 8 y - 24 = 0$
$2 g = - 6, 2 f = - 8, c = - 24$
$g = - 3, f = - 4, c = - 24$
$(- g, - f) = (3, 4) r_{2} = \sqrt{(+ 3)^{2} + (4)^{2} - (- 24)}$
$= \sqrt{9 + 16 + 24} = \sqrt{49} = 7$
$c_{2} = (3, 4), r_{2} = 7$
$c_{1} c_{2} = \sqrt{(0 - 3)^{2} + (0 - 4)^{2}}$
$= \sqrt{9 + 16} = \sqrt{25} = 5$
$∴ c_{1} c_{2} = r_{2} - r_{1}$
$5 = 7 - 2 = 5$
$5 = 5$
Hence the given circles touch each other internally. Hence the number of common tangent is only one

BCECE-2012

Conic Section

119862 The circle $S_{1}$ with centre $C_{1} (a_{1}, b_{1})$ and radius $r_{1}$ touches externally the circle $S_{2}$ with centre $C_{2} (a_{2}, b_{2})$ and radius $r_{2}$ . If the tangent at their common point passes through the origin, then

1

(a_{1}^{2} + a_{2}^{2}) + (b_{1}^{2} + b_{2}^{2}) = r_{1}^{2} + r_{2}^{2}

2

(a_{1}^{2} - a_{2}^{2}) + (b_{1}^{2} - b_{2}^{2}) = r_{1}^{2} - r_{2}^{2}

3

{(a_{1}^{2} - b_{2})}^{2} + (a_{2}^{2} + b_{2}^{2}) = r_{1}^{2} + r_{2}^{2}

4

(a_{1}^{2} - b_{1}^{2}) + (a_{1}^{2} + b_{2}^{2}) = r_{1}^{2} + r_{2}^{2}

Explanation:

B According to question we have
$S_{1} = {(x - a_{1})}^{2} + {(y - b_{1})}^{2} = r_{1}^{2}$
$S_{2} = {(x - a_{2})}^{2} + {(y - b_{2})}^{2} = r_{2}^{2}$
$for common tangent,$
$S_{1} - S_{2} = 0$
$2 x (a_{1} - a_{2}) + 2 y (b_{1} - b_{2}) + (a_{2}^{2} + b_{2}^{2}) - (a_{1}^{2} + b_{1}^{2}) + r_{1}^{2}$
$- r_{2}^{2} = 0$
$if this passes through the origin then$
$(a_{2}^{2} + b_{2}^{2}) - (a_{1}^{2} + b_{1}^{2}) + r_{1}^{2} - r_{2}^{2} = 0$
$(a_{2}^{2} - a_{1}^{2}) + (b_{2}^{2} - b_{1}^{2}) = r_{2}^{2} - r_{1}^{2}$
${(a_{1}^{2} - a_{2})}^{2}) + (b_{1}^{2} - b_{2}^{2}) = (r_{1}^{2} - r_{2}^{2})$

BCECE-2011

Conic Section

119863 The equation of a circle of radius 5 which lies within the circle $x^{2} + y^{2} + 14 x + 10 y - 26 = 0$ and touches it at the point $(- 1, 3)$ is

1

x^{2} + y^{2} + 8 x + 2 y + 8 = 0

2

x^{2} + y^{2} + 8 x + 2 y - 8 = 0

3

x^{2} + y^{2} + 8 x + 2 y - 14 = 0

4 None of the above

Explanation:

B $x^{2} + y^{2} + 14 x + 10 y - 26 = 0$
$2 g = 14, 2 f = 10, c = - 26$
$g = 7, f = + 5, c = - 26$
$(- g, - f) = (- 7, - 5), r = \sqrt{(7)^{2} + (5)^{2} - (- 26)}$
$= \sqrt{49 + 25 + 26}$
$= \sqrt{100} = 10$
$(- 4, - 10)$
$r = 5$
According to question, other circle lies in the circle touching it at $(- 1, 3)$ and its radius is 5 , So it also passes through the centre of the given circle which is $(- 7, - 5)$ Now we have the ends point of diameter of the circle
So, its centre is $(\frac{- 7 - 1}{2}, \frac{- 5 + 3}{2})$
$r = 5$
So the equation of the other circle is $(x + 4)^{2} + (y + 1)^{2} = 5^{2}$ $x^{2} + 16 + 8 x + y^{2} + 1 + 2 y = 25$ $x^{2} + y^{2} + 8 x + 2 y - 8 = 0$

BCECE-2010

Conic Section

119864 The number of common tangents to two circles $x^{2} + y^{2} = 4$ and $x^{2} + y^{2} - 8 x + 12 = 0$ is

1 1

2 2

3 3

4 4

Explanation:

C Given equations of circles are
$x^{2} + y^{2} = 4$
and $x^{2} + y^{2} - 8 x + 12 = 0$
or
$x^{2} + y^{2} = (2)^{2}$
$(x - 4)^{2} + (y - 0)^{2} = (2)^{2}$
The figure of both the circles are shown blow:
original image
From figure, we see that there is exactly three common tangents.

CG PET- 2015

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation: