QBManager

Conic Section

119989 For the given circles $x^{2} + y^{2} - 6 x - 2 y + 1 = 0$ and $x^{2} + y^{2} + 2 x - 8 y + 13 = 0$ , which of the following is true?

1 One circle lies inside the other

2 One circle lies completely outside the other

3 Two circle intersect in two points

4 They touch each other externally

Explanation:

D : Given,
$x^{2} + y^{2} - 6 x - 2 y + 1 = 0$
$x^{2} - 6 x + 9 - 9 + y^{2} - 2 y + 1 = 0$
$(x - 3)^{2} + (y - 1)^{2} = 9$
$C_{1} = (3, 1), S_{1} = 3$
$x^{2} + y^{2} + 2 x - 8 y + 13 = 0$
$x^{2} + 2 x + 1 - 1 + y^{2} - 8 y + 16 - 16 + 13 = 0$
$(x + 1)^{2} + (y - 4)^{2} = 4$
$C_{2} = (- 1, 4)$ and $S_{2} = 2$
Here
$∴ C_{1} C_{2} = \sqrt{(3 - (- 1))^{2} + ((+ 1) - 4)^{2}} = \sqrt{16 + 9} = \sqrt{25}$
$C_{1} C_{2} = 5, S_{1} + S_{2} = 3 + 2 = 5$
$∴ C_{1} C_{2} = S_{1} + S_{2}$
Hence two circles touch externally.

BCECE-2016

Conic Section

119990 The circles $x^{2} + y^{2} - 5 x + 6 y + 15 = 0$ and $x^{2} + y^{2} - 2 x + 6 y + 6 = 0$ touch each other

1 internally

2 externally

3 Do not say anything

4 None of these

Explanation:

A Given,
$x^{2} + y^{2} - 5 x + 6 y + 15 = 0$
$x^{2} + y^{2} - 2 x + 6 y + 6 = 0$
From equation (i), we get -
$x^{2} + y^{2} - 5 x + 6 y + 15 = 0$
$2 g_{1} = - 5, 2 f_{1} = 6, C_{1} = 15$
$g_{1} = \frac{- 5}{2}, f_{1} = 3, C_{1} = 15, (- g_{1}, - f_{1}) \to C_{1} = (\frac{5}{2}, - 3)$
$S_{1} = \sqrt{g_{1}^{2} + f_{1}^{2} - C_{1}} = \sqrt{{(\frac{- 5}{2})}^{2} + 3^{2} - 15} = \sqrt{\frac{25}{4} + 9 - 15}$
$= \sqrt{\frac{25}{4} - 6} = \sqrt{\frac{1}{4}} = \frac{1}{2}$
From equation (ii), we get -
$x^{2} + y^{2} - 2 x + 6 y + 6 = 0$
$2 g_{2} = - 2, 2 f_{2} = 6, C_{2} = 6$
$g_{2} = - 1, f_{2} = 3, C_{2} = 6, C_{2} \to (- g_{2}, f_{2}) = (1, - 3)$
$S_{2} = \sqrt{g_{2}^{2} + f_{2}^{2} - C_{2}} = \sqrt{(- 1)^{2} + (3)^{2} - 6} = \sqrt{1 + 9 - 6} = \sqrt{4} = 2$
$∴ C_{1} C_{2} = \sqrt{{(\frac{5}{2} - 1)}^{2} + (- 3 - (- 3))^{2}} = \sqrt{\frac{9}{4} + 0}$
$C_{1} C_{2} = \frac{3}{2}$
$∴ C_{1} C_{2} = | S_{1} - S_{2} |$
$\frac{3}{2} = | \frac{1}{2} - 2 | = \frac{3}{2}$
since
Hence, the two circle touch each other internally.

BCECE-2014

Conic Section

119991 If two circles $(x - 1)^{2} + (y - 3)^{2} = r^{2}$ and $x^{2} + y^{2} - 8 x + 2 y + 8 = 0$ intersect in two distinct points, then

1

2 < r < 8

2

r < 2

3

r = 2

4

r > 2

Explanation:

A Given,Two circle $(x - 1)^{2} + (y - 3)^{2} = r^{2}$
And $x^{2} + y^{2} - 8 x + 2 y + 8 = 0$
Equation of first circle,
$(x - 1)^{2} + (y - 3)^{2} = r^{2}$
$Center C_{1} = (1, 3), radius r_{1} = r$
Equation of second circle,
$x^{2} + y^{2} - 8 x + 2 y + 8 = 0$
centre $C_{2} = (4, - 1)$
and radius $r_{2} = \sqrt{4^{2} + 1^{2} - 8} = \sqrt{17 - 8} = 3$
Two circle intersect in two distinct points, then
$r_{1} - r_{2} < C_{1} C_{2} < r_{1} + r_{2}$
$r - 3 < \sqrt{(4 - 1)^{2} + (- 1 - 3)^{2}} < r + 3$
$r - 3 < \sqrt{9 + 16} < r + 3$
$r - 3 < 5 < r + 3$
$r < 8 and 2 < r$
$2 < r < 8$

AP EAMCET-20.04.2019

Conic Section

119992 The locus of the point of intersection of the tangents at the extremities of a chord of the circle $x^{2} + y^{2} = a^{2}$ which touches the circle $x^{2} + y^{2} - 2 a x = 0$ passes through the point

1

(a / 2, 0)

2

(0, a / 2)

3

(a, 0)

4

(0, 0)

Explanation:

A
Let the point of intersection of tangents be $(h, k)$
Then the chord of contact
$xh + yk = a^{2}$
The chord touched the circle
$x^{2} + y^{2} - 2 ax = 0$
centre $(a, 0)$ and Radius $= a$
Thus perpendicular from the centre on line is equal to radius of circle. Thus we get
$\sqrt{h^{2} + k^{2}} = \pm a$
$ah - a^{2} = a \sqrt{h^{2} + k^{2}}$
$h - a = \sqrt{h^{2} + k^{2}}$
Squaring both side we get
$(h - a)^{2} = h^{2} + k^{2}$
$h^{2} + a^{2} - 2 a h = h^{2} + k^{2}$
$k^{2} + 2 a h = a^{2}$
Locus of the point $y^{2} + 2 a x = a^{2}$
Hence, the curve passes through the point $(\frac{a}{2}, 0)$

AMU-2011

Conic Section

119993 The angle between circles $x^{2} + y^{2} + 2 x + 4 y + 1$ $= 0$ and $x^{2} + y^{2} - 2 x + 6 y - 3 = 0$ is:

1

\cos^{- 1} (\frac{3}{\sqrt{13}})

2

\cos^{- 1} (\frac{3}{\sqrt{31}})

3

\cos^{- 1} (\sqrt{\frac{3}{31}})

4

2 \cos^{- 1} (\frac{3}{\sqrt{13}})

Explanation:

A : Given,
$C_{1} : x^{2} + y^{2} + 2 x + 4 y + 1 = 0$
$C_{2} : x^{2} + y^{2} - 2 x + 6 y - 3 = 0$
$∴ C_{1} = (- 1, - 2) and r_{1} = \sqrt{1 + 4 - 1} = 2$
$C_{2} = (1, - 3) and r_{2} = \sqrt{1 + 9 + 3} = \sqrt{13}$
$∴ d = C_{1} C_{2} = \sqrt{(- 1 - 1)^{2} + (- 2 + 3)^{2}} = \sqrt{4 + 1} = \sqrt{5}$
$∴ \cos θ = | \frac{d^{2} - r_{1}^{2} - r_{2}^{2}}{2 r_{1} r_{2}} | = | \frac{5 - 4 - 13}{2 \times 2 \times \sqrt{13}} | = | \frac{- 3}{\sqrt{13}} |$
$∴ θ = \cos^{- 1} (\frac{3}{\sqrt{13}})$

APEAPCET-20.08.2021

Conic Section

119989 For the given circles $x^{2} + y^{2} - 6 x - 2 y + 1 = 0$ and $x^{2} + y^{2} + 2 x - 8 y + 13 = 0$ , which of the following is true?

1 One circle lies inside the other

2 One circle lies completely outside the other

3 Two circle intersect in two points

4 They touch each other externally

Explanation:

D : Given,
$x^{2} + y^{2} - 6 x - 2 y + 1 = 0$
$x^{2} - 6 x + 9 - 9 + y^{2} - 2 y + 1 = 0$
$(x - 3)^{2} + (y - 1)^{2} = 9$
$C_{1} = (3, 1), S_{1} = 3$
$x^{2} + y^{2} + 2 x - 8 y + 13 = 0$
$x^{2} + 2 x + 1 - 1 + y^{2} - 8 y + 16 - 16 + 13 = 0$
$(x + 1)^{2} + (y - 4)^{2} = 4$
$C_{2} = (- 1, 4)$ and $S_{2} = 2$
Here
$∴ C_{1} C_{2} = \sqrt{(3 - (- 1))^{2} + ((+ 1) - 4)^{2}} = \sqrt{16 + 9} = \sqrt{25}$
$C_{1} C_{2} = 5, S_{1} + S_{2} = 3 + 2 = 5$
$∴ C_{1} C_{2} = S_{1} + S_{2}$
Hence two circles touch externally.

BCECE-2016

Conic Section

119990 The circles $x^{2} + y^{2} - 5 x + 6 y + 15 = 0$ and $x^{2} + y^{2} - 2 x + 6 y + 6 = 0$ touch each other

1 internally

2 externally

3 Do not say anything

4 None of these

Explanation:

A Given,
$x^{2} + y^{2} - 5 x + 6 y + 15 = 0$
$x^{2} + y^{2} - 2 x + 6 y + 6 = 0$
From equation (i), we get -
$x^{2} + y^{2} - 5 x + 6 y + 15 = 0$
$2 g_{1} = - 5, 2 f_{1} = 6, C_{1} = 15$
$g_{1} = \frac{- 5}{2}, f_{1} = 3, C_{1} = 15, (- g_{1}, - f_{1}) \to C_{1} = (\frac{5}{2}, - 3)$
$S_{1} = \sqrt{g_{1}^{2} + f_{1}^{2} - C_{1}} = \sqrt{{(\frac{- 5}{2})}^{2} + 3^{2} - 15} = \sqrt{\frac{25}{4} + 9 - 15}$
$= \sqrt{\frac{25}{4} - 6} = \sqrt{\frac{1}{4}} = \frac{1}{2}$
From equation (ii), we get -
$x^{2} + y^{2} - 2 x + 6 y + 6 = 0$
$2 g_{2} = - 2, 2 f_{2} = 6, C_{2} = 6$
$g_{2} = - 1, f_{2} = 3, C_{2} = 6, C_{2} \to (- g_{2}, f_{2}) = (1, - 3)$
$S_{2} = \sqrt{g_{2}^{2} + f_{2}^{2} - C_{2}} = \sqrt{(- 1)^{2} + (3)^{2} - 6} = \sqrt{1 + 9 - 6} = \sqrt{4} = 2$
$∴ C_{1} C_{2} = \sqrt{{(\frac{5}{2} - 1)}^{2} + (- 3 - (- 3))^{2}} = \sqrt{\frac{9}{4} + 0}$
$C_{1} C_{2} = \frac{3}{2}$
$∴ C_{1} C_{2} = | S_{1} - S_{2} |$
$\frac{3}{2} = | \frac{1}{2} - 2 | = \frac{3}{2}$
since
Hence, the two circle touch each other internally.

BCECE-2014

Conic Section

119991 If two circles $(x - 1)^{2} + (y - 3)^{2} = r^{2}$ and $x^{2} + y^{2} - 8 x + 2 y + 8 = 0$ intersect in two distinct points, then

1

2 < r < 8

2

r < 2

3

r = 2

4

r > 2

Explanation:

A Given,Two circle $(x - 1)^{2} + (y - 3)^{2} = r^{2}$
And $x^{2} + y^{2} - 8 x + 2 y + 8 = 0$
Equation of first circle,
$(x - 1)^{2} + (y - 3)^{2} = r^{2}$
$Center C_{1} = (1, 3), radius r_{1} = r$
Equation of second circle,
$x^{2} + y^{2} - 8 x + 2 y + 8 = 0$
centre $C_{2} = (4, - 1)$
and radius $r_{2} = \sqrt{4^{2} + 1^{2} - 8} = \sqrt{17 - 8} = 3$
Two circle intersect in two distinct points, then
$r_{1} - r_{2} < C_{1} C_{2} < r_{1} + r_{2}$
$r - 3 < \sqrt{(4 - 1)^{2} + (- 1 - 3)^{2}} < r + 3$
$r - 3 < \sqrt{9 + 16} < r + 3$
$r - 3 < 5 < r + 3$
$r < 8 and 2 < r$
$2 < r < 8$

AP EAMCET-20.04.2019

Conic Section

119992 The locus of the point of intersection of the tangents at the extremities of a chord of the circle $x^{2} + y^{2} = a^{2}$ which touches the circle $x^{2} + y^{2} - 2 a x = 0$ passes through the point

1

(a / 2, 0)

2

(0, a / 2)

3

(a, 0)

4

(0, 0)

Explanation:

A
Let the point of intersection of tangents be $(h, k)$
Then the chord of contact
$xh + yk = a^{2}$
The chord touched the circle
$x^{2} + y^{2} - 2 ax = 0$
centre $(a, 0)$ and Radius $= a$
Thus perpendicular from the centre on line is equal to radius of circle. Thus we get
$\sqrt{h^{2} + k^{2}} = \pm a$
$ah - a^{2} = a \sqrt{h^{2} + k^{2}}$
$h - a = \sqrt{h^{2} + k^{2}}$
Squaring both side we get
$(h - a)^{2} = h^{2} + k^{2}$
$h^{2} + a^{2} - 2 a h = h^{2} + k^{2}$
$k^{2} + 2 a h = a^{2}$
Locus of the point $y^{2} + 2 a x = a^{2}$
Hence, the curve passes through the point $(\frac{a}{2}, 0)$

AMU-2011

Conic Section

119993 The angle between circles $x^{2} + y^{2} + 2 x + 4 y + 1$ $= 0$ and $x^{2} + y^{2} - 2 x + 6 y - 3 = 0$ is:

1

\cos^{- 1} (\frac{3}{\sqrt{13}})

2

\cos^{- 1} (\frac{3}{\sqrt{31}})

3

\cos^{- 1} (\sqrt{\frac{3}{31}})

4

2 \cos^{- 1} (\frac{3}{\sqrt{13}})

Explanation:

A : Given,
$C_{1} : x^{2} + y^{2} + 2 x + 4 y + 1 = 0$
$C_{2} : x^{2} + y^{2} - 2 x + 6 y - 3 = 0$
$∴ C_{1} = (- 1, - 2) and r_{1} = \sqrt{1 + 4 - 1} = 2$
$C_{2} = (1, - 3) and r_{2} = \sqrt{1 + 9 + 3} = \sqrt{13}$
$∴ d = C_{1} C_{2} = \sqrt{(- 1 - 1)^{2} + (- 2 + 3)^{2}} = \sqrt{4 + 1} = \sqrt{5}$
$∴ \cos θ = | \frac{d^{2} - r_{1}^{2} - r_{2}^{2}}{2 r_{1} r_{2}} | = | \frac{5 - 4 - 13}{2 \times 2 \times \sqrt{13}} | = | \frac{- 3}{\sqrt{13}} |$
$∴ θ = \cos^{- 1} (\frac{3}{\sqrt{13}})$

APEAPCET-20.08.2021

Conic Section

119989 For the given circles $x^{2} + y^{2} - 6 x - 2 y + 1 = 0$ and $x^{2} + y^{2} + 2 x - 8 y + 13 = 0$ , which of the following is true?

1 One circle lies inside the other

2 One circle lies completely outside the other

3 Two circle intersect in two points

4 They touch each other externally

Explanation:

D : Given,
$x^{2} + y^{2} - 6 x - 2 y + 1 = 0$
$x^{2} - 6 x + 9 - 9 + y^{2} - 2 y + 1 = 0$
$(x - 3)^{2} + (y - 1)^{2} = 9$
$C_{1} = (3, 1), S_{1} = 3$
$x^{2} + y^{2} + 2 x - 8 y + 13 = 0$
$x^{2} + 2 x + 1 - 1 + y^{2} - 8 y + 16 - 16 + 13 = 0$
$(x + 1)^{2} + (y - 4)^{2} = 4$
$C_{2} = (- 1, 4)$ and $S_{2} = 2$
Here
$∴ C_{1} C_{2} = \sqrt{(3 - (- 1))^{2} + ((+ 1) - 4)^{2}} = \sqrt{16 + 9} = \sqrt{25}$
$C_{1} C_{2} = 5, S_{1} + S_{2} = 3 + 2 = 5$
$∴ C_{1} C_{2} = S_{1} + S_{2}$
Hence two circles touch externally.

BCECE-2016

Conic Section

119990 The circles $x^{2} + y^{2} - 5 x + 6 y + 15 = 0$ and $x^{2} + y^{2} - 2 x + 6 y + 6 = 0$ touch each other

1 internally

2 externally

3 Do not say anything

4 None of these

Explanation:

A Given,
$x^{2} + y^{2} - 5 x + 6 y + 15 = 0$
$x^{2} + y^{2} - 2 x + 6 y + 6 = 0$
From equation (i), we get -
$x^{2} + y^{2} - 5 x + 6 y + 15 = 0$
$2 g_{1} = - 5, 2 f_{1} = 6, C_{1} = 15$
$g_{1} = \frac{- 5}{2}, f_{1} = 3, C_{1} = 15, (- g_{1}, - f_{1}) \to C_{1} = (\frac{5}{2}, - 3)$
$S_{1} = \sqrt{g_{1}^{2} + f_{1}^{2} - C_{1}} = \sqrt{{(\frac{- 5}{2})}^{2} + 3^{2} - 15} = \sqrt{\frac{25}{4} + 9 - 15}$
$= \sqrt{\frac{25}{4} - 6} = \sqrt{\frac{1}{4}} = \frac{1}{2}$
From equation (ii), we get -
$x^{2} + y^{2} - 2 x + 6 y + 6 = 0$
$2 g_{2} = - 2, 2 f_{2} = 6, C_{2} = 6$
$g_{2} = - 1, f_{2} = 3, C_{2} = 6, C_{2} \to (- g_{2}, f_{2}) = (1, - 3)$
$S_{2} = \sqrt{g_{2}^{2} + f_{2}^{2} - C_{2}} = \sqrt{(- 1)^{2} + (3)^{2} - 6} = \sqrt{1 + 9 - 6} = \sqrt{4} = 2$
$∴ C_{1} C_{2} = \sqrt{{(\frac{5}{2} - 1)}^{2} + (- 3 - (- 3))^{2}} = \sqrt{\frac{9}{4} + 0}$
$C_{1} C_{2} = \frac{3}{2}$
$∴ C_{1} C_{2} = | S_{1} - S_{2} |$
$\frac{3}{2} = | \frac{1}{2} - 2 | = \frac{3}{2}$
since
Hence, the two circle touch each other internally.

BCECE-2014

Conic Section

119991 If two circles $(x - 1)^{2} + (y - 3)^{2} = r^{2}$ and $x^{2} + y^{2} - 8 x + 2 y + 8 = 0$ intersect in two distinct points, then

1

2 < r < 8

2

r < 2

3

r = 2

4

r > 2

Explanation:

A Given,Two circle $(x - 1)^{2} + (y - 3)^{2} = r^{2}$
And $x^{2} + y^{2} - 8 x + 2 y + 8 = 0$
Equation of first circle,
$(x - 1)^{2} + (y - 3)^{2} = r^{2}$
$Center C_{1} = (1, 3), radius r_{1} = r$
Equation of second circle,
$x^{2} + y^{2} - 8 x + 2 y + 8 = 0$
centre $C_{2} = (4, - 1)$
and radius $r_{2} = \sqrt{4^{2} + 1^{2} - 8} = \sqrt{17 - 8} = 3$
Two circle intersect in two distinct points, then
$r_{1} - r_{2} < C_{1} C_{2} < r_{1} + r_{2}$
$r - 3 < \sqrt{(4 - 1)^{2} + (- 1 - 3)^{2}} < r + 3$
$r - 3 < \sqrt{9 + 16} < r + 3$
$r - 3 < 5 < r + 3$
$r < 8 and 2 < r$
$2 < r < 8$

AP EAMCET-20.04.2019

Conic Section

119992 The locus of the point of intersection of the tangents at the extremities of a chord of the circle $x^{2} + y^{2} = a^{2}$ which touches the circle $x^{2} + y^{2} - 2 a x = 0$ passes through the point

1

(a / 2, 0)

2

(0, a / 2)

3

(a, 0)

4

(0, 0)

Explanation:

A
Let the point of intersection of tangents be $(h, k)$
Then the chord of contact
$xh + yk = a^{2}$
The chord touched the circle
$x^{2} + y^{2} - 2 ax = 0$
centre $(a, 0)$ and Radius $= a$
Thus perpendicular from the centre on line is equal to radius of circle. Thus we get
$\sqrt{h^{2} + k^{2}} = \pm a$
$ah - a^{2} = a \sqrt{h^{2} + k^{2}}$
$h - a = \sqrt{h^{2} + k^{2}}$
Squaring both side we get
$(h - a)^{2} = h^{2} + k^{2}$
$h^{2} + a^{2} - 2 a h = h^{2} + k^{2}$
$k^{2} + 2 a h = a^{2}$
Locus of the point $y^{2} + 2 a x = a^{2}$
Hence, the curve passes through the point $(\frac{a}{2}, 0)$

AMU-2011

Conic Section

119993 The angle between circles $x^{2} + y^{2} + 2 x + 4 y + 1$ $= 0$ and $x^{2} + y^{2} - 2 x + 6 y - 3 = 0$ is:

1

\cos^{- 1} (\frac{3}{\sqrt{13}})

2

\cos^{- 1} (\frac{3}{\sqrt{31}})

3

\cos^{- 1} (\sqrt{\frac{3}{31}})

4

2 \cos^{- 1} (\frac{3}{\sqrt{13}})

Explanation:

A : Given,
$C_{1} : x^{2} + y^{2} + 2 x + 4 y + 1 = 0$
$C_{2} : x^{2} + y^{2} - 2 x + 6 y - 3 = 0$
$∴ C_{1} = (- 1, - 2) and r_{1} = \sqrt{1 + 4 - 1} = 2$
$C_{2} = (1, - 3) and r_{2} = \sqrt{1 + 9 + 3} = \sqrt{13}$
$∴ d = C_{1} C_{2} = \sqrt{(- 1 - 1)^{2} + (- 2 + 3)^{2}} = \sqrt{4 + 1} = \sqrt{5}$
$∴ \cos θ = | \frac{d^{2} - r_{1}^{2} - r_{2}^{2}}{2 r_{1} r_{2}} | = | \frac{5 - 4 - 13}{2 \times 2 \times \sqrt{13}} | = | \frac{- 3}{\sqrt{13}} |$
$∴ θ = \cos^{- 1} (\frac{3}{\sqrt{13}})$

APEAPCET-20.08.2021

Conic Section

119989 For the given circles $x^{2} + y^{2} - 6 x - 2 y + 1 = 0$ and $x^{2} + y^{2} + 2 x - 8 y + 13 = 0$ , which of the following is true?

1 One circle lies inside the other

2 One circle lies completely outside the other

3 Two circle intersect in two points

4 They touch each other externally

Explanation:

D : Given,
$x^{2} + y^{2} - 6 x - 2 y + 1 = 0$
$x^{2} - 6 x + 9 - 9 + y^{2} - 2 y + 1 = 0$
$(x - 3)^{2} + (y - 1)^{2} = 9$
$C_{1} = (3, 1), S_{1} = 3$
$x^{2} + y^{2} + 2 x - 8 y + 13 = 0$
$x^{2} + 2 x + 1 - 1 + y^{2} - 8 y + 16 - 16 + 13 = 0$
$(x + 1)^{2} + (y - 4)^{2} = 4$
$C_{2} = (- 1, 4)$ and $S_{2} = 2$
Here
$∴ C_{1} C_{2} = \sqrt{(3 - (- 1))^{2} + ((+ 1) - 4)^{2}} = \sqrt{16 + 9} = \sqrt{25}$
$C_{1} C_{2} = 5, S_{1} + S_{2} = 3 + 2 = 5$
$∴ C_{1} C_{2} = S_{1} + S_{2}$
Hence two circles touch externally.

BCECE-2016

Conic Section

119990 The circles $x^{2} + y^{2} - 5 x + 6 y + 15 = 0$ and $x^{2} + y^{2} - 2 x + 6 y + 6 = 0$ touch each other

1 internally

2 externally

3 Do not say anything

4 None of these

Explanation:

A Given,
$x^{2} + y^{2} - 5 x + 6 y + 15 = 0$
$x^{2} + y^{2} - 2 x + 6 y + 6 = 0$
From equation (i), we get -
$x^{2} + y^{2} - 5 x + 6 y + 15 = 0$
$2 g_{1} = - 5, 2 f_{1} = 6, C_{1} = 15$
$g_{1} = \frac{- 5}{2}, f_{1} = 3, C_{1} = 15, (- g_{1}, - f_{1}) \to C_{1} = (\frac{5}{2}, - 3)$
$S_{1} = \sqrt{g_{1}^{2} + f_{1}^{2} - C_{1}} = \sqrt{{(\frac{- 5}{2})}^{2} + 3^{2} - 15} = \sqrt{\frac{25}{4} + 9 - 15}$
$= \sqrt{\frac{25}{4} - 6} = \sqrt{\frac{1}{4}} = \frac{1}{2}$
From equation (ii), we get -
$x^{2} + y^{2} - 2 x + 6 y + 6 = 0$
$2 g_{2} = - 2, 2 f_{2} = 6, C_{2} = 6$
$g_{2} = - 1, f_{2} = 3, C_{2} = 6, C_{2} \to (- g_{2}, f_{2}) = (1, - 3)$
$S_{2} = \sqrt{g_{2}^{2} + f_{2}^{2} - C_{2}} = \sqrt{(- 1)^{2} + (3)^{2} - 6} = \sqrt{1 + 9 - 6} = \sqrt{4} = 2$
$∴ C_{1} C_{2} = \sqrt{{(\frac{5}{2} - 1)}^{2} + (- 3 - (- 3))^{2}} = \sqrt{\frac{9}{4} + 0}$
$C_{1} C_{2} = \frac{3}{2}$
$∴ C_{1} C_{2} = | S_{1} - S_{2} |$
$\frac{3}{2} = | \frac{1}{2} - 2 | = \frac{3}{2}$
since
Hence, the two circle touch each other internally.

BCECE-2014

Conic Section

119991 If two circles $(x - 1)^{2} + (y - 3)^{2} = r^{2}$ and $x^{2} + y^{2} - 8 x + 2 y + 8 = 0$ intersect in two distinct points, then

1

2 < r < 8

2

r < 2

3

r = 2

4

r > 2

Explanation:

A Given,Two circle $(x - 1)^{2} + (y - 3)^{2} = r^{2}$
And $x^{2} + y^{2} - 8 x + 2 y + 8 = 0$
Equation of first circle,
$(x - 1)^{2} + (y - 3)^{2} = r^{2}$
$Center C_{1} = (1, 3), radius r_{1} = r$
Equation of second circle,
$x^{2} + y^{2} - 8 x + 2 y + 8 = 0$
centre $C_{2} = (4, - 1)$
and radius $r_{2} = \sqrt{4^{2} + 1^{2} - 8} = \sqrt{17 - 8} = 3$
Two circle intersect in two distinct points, then
$r_{1} - r_{2} < C_{1} C_{2} < r_{1} + r_{2}$
$r - 3 < \sqrt{(4 - 1)^{2} + (- 1 - 3)^{2}} < r + 3$
$r - 3 < \sqrt{9 + 16} < r + 3$
$r - 3 < 5 < r + 3$
$r < 8 and 2 < r$
$2 < r < 8$

AP EAMCET-20.04.2019

Conic Section

119992 The locus of the point of intersection of the tangents at the extremities of a chord of the circle $x^{2} + y^{2} = a^{2}$ which touches the circle $x^{2} + y^{2} - 2 a x = 0$ passes through the point

1

(a / 2, 0)

2

(0, a / 2)

3

(a, 0)

4

(0, 0)

Explanation:

A
Let the point of intersection of tangents be $(h, k)$
Then the chord of contact
$xh + yk = a^{2}$
The chord touched the circle
$x^{2} + y^{2} - 2 ax = 0$
centre $(a, 0)$ and Radius $= a$
Thus perpendicular from the centre on line is equal to radius of circle. Thus we get
$\sqrt{h^{2} + k^{2}} = \pm a$
$ah - a^{2} = a \sqrt{h^{2} + k^{2}}$
$h - a = \sqrt{h^{2} + k^{2}}$
Squaring both side we get
$(h - a)^{2} = h^{2} + k^{2}$
$h^{2} + a^{2} - 2 a h = h^{2} + k^{2}$
$k^{2} + 2 a h = a^{2}$
Locus of the point $y^{2} + 2 a x = a^{2}$
Hence, the curve passes through the point $(\frac{a}{2}, 0)$

AMU-2011

Conic Section

119993 The angle between circles $x^{2} + y^{2} + 2 x + 4 y + 1$ $= 0$ and $x^{2} + y^{2} - 2 x + 6 y - 3 = 0$ is:

1

\cos^{- 1} (\frac{3}{\sqrt{13}})

2

\cos^{- 1} (\frac{3}{\sqrt{31}})

3

\cos^{- 1} (\sqrt{\frac{3}{31}})

4

2 \cos^{- 1} (\frac{3}{\sqrt{13}})

Explanation:

A : Given,
$C_{1} : x^{2} + y^{2} + 2 x + 4 y + 1 = 0$
$C_{2} : x^{2} + y^{2} - 2 x + 6 y - 3 = 0$
$∴ C_{1} = (- 1, - 2) and r_{1} = \sqrt{1 + 4 - 1} = 2$
$C_{2} = (1, - 3) and r_{2} = \sqrt{1 + 9 + 3} = \sqrt{13}$
$∴ d = C_{1} C_{2} = \sqrt{(- 1 - 1)^{2} + (- 2 + 3)^{2}} = \sqrt{4 + 1} = \sqrt{5}$
$∴ \cos θ = | \frac{d^{2} - r_{1}^{2} - r_{2}^{2}}{2 r_{1} r_{2}} | = | \frac{5 - 4 - 13}{2 \times 2 \times \sqrt{13}} | = | \frac{- 3}{\sqrt{13}} |$
$∴ θ = \cos^{- 1} (\frac{3}{\sqrt{13}})$

APEAPCET-20.08.2021

Conic Section

119989 For the given circles $x^{2} + y^{2} - 6 x - 2 y + 1 = 0$ and $x^{2} + y^{2} + 2 x - 8 y + 13 = 0$ , which of the following is true?

1 One circle lies inside the other

2 One circle lies completely outside the other

3 Two circle intersect in two points

4 They touch each other externally

Explanation:

D : Given,
$x^{2} + y^{2} - 6 x - 2 y + 1 = 0$
$x^{2} - 6 x + 9 - 9 + y^{2} - 2 y + 1 = 0$
$(x - 3)^{2} + (y - 1)^{2} = 9$
$C_{1} = (3, 1), S_{1} = 3$
$x^{2} + y^{2} + 2 x - 8 y + 13 = 0$
$x^{2} + 2 x + 1 - 1 + y^{2} - 8 y + 16 - 16 + 13 = 0$
$(x + 1)^{2} + (y - 4)^{2} = 4$
$C_{2} = (- 1, 4)$ and $S_{2} = 2$
Here
$∴ C_{1} C_{2} = \sqrt{(3 - (- 1))^{2} + ((+ 1) - 4)^{2}} = \sqrt{16 + 9} = \sqrt{25}$
$C_{1} C_{2} = 5, S_{1} + S_{2} = 3 + 2 = 5$
$∴ C_{1} C_{2} = S_{1} + S_{2}$
Hence two circles touch externally.

BCECE-2016

Conic Section

119990 The circles $x^{2} + y^{2} - 5 x + 6 y + 15 = 0$ and $x^{2} + y^{2} - 2 x + 6 y + 6 = 0$ touch each other

1 internally

2 externally

3 Do not say anything

4 None of these

Explanation:

A Given,
$x^{2} + y^{2} - 5 x + 6 y + 15 = 0$
$x^{2} + y^{2} - 2 x + 6 y + 6 = 0$
From equation (i), we get -
$x^{2} + y^{2} - 5 x + 6 y + 15 = 0$
$2 g_{1} = - 5, 2 f_{1} = 6, C_{1} = 15$
$g_{1} = \frac{- 5}{2}, f_{1} = 3, C_{1} = 15, (- g_{1}, - f_{1}) \to C_{1} = (\frac{5}{2}, - 3)$
$S_{1} = \sqrt{g_{1}^{2} + f_{1}^{2} - C_{1}} = \sqrt{{(\frac{- 5}{2})}^{2} + 3^{2} - 15} = \sqrt{\frac{25}{4} + 9 - 15}$
$= \sqrt{\frac{25}{4} - 6} = \sqrt{\frac{1}{4}} = \frac{1}{2}$
From equation (ii), we get -
$x^{2} + y^{2} - 2 x + 6 y + 6 = 0$
$2 g_{2} = - 2, 2 f_{2} = 6, C_{2} = 6$
$g_{2} = - 1, f_{2} = 3, C_{2} = 6, C_{2} \to (- g_{2}, f_{2}) = (1, - 3)$
$S_{2} = \sqrt{g_{2}^{2} + f_{2}^{2} - C_{2}} = \sqrt{(- 1)^{2} + (3)^{2} - 6} = \sqrt{1 + 9 - 6} = \sqrt{4} = 2$
$∴ C_{1} C_{2} = \sqrt{{(\frac{5}{2} - 1)}^{2} + (- 3 - (- 3))^{2}} = \sqrt{\frac{9}{4} + 0}$
$C_{1} C_{2} = \frac{3}{2}$
$∴ C_{1} C_{2} = | S_{1} - S_{2} |$
$\frac{3}{2} = | \frac{1}{2} - 2 | = \frac{3}{2}$
since
Hence, the two circle touch each other internally.

BCECE-2014

Conic Section

119991 If two circles $(x - 1)^{2} + (y - 3)^{2} = r^{2}$ and $x^{2} + y^{2} - 8 x + 2 y + 8 = 0$ intersect in two distinct points, then

1

2 < r < 8

2

r < 2

3

r = 2

4

r > 2

Explanation:

A Given,Two circle $(x - 1)^{2} + (y - 3)^{2} = r^{2}$
And $x^{2} + y^{2} - 8 x + 2 y + 8 = 0$
Equation of first circle,
$(x - 1)^{2} + (y - 3)^{2} = r^{2}$
$Center C_{1} = (1, 3), radius r_{1} = r$
Equation of second circle,
$x^{2} + y^{2} - 8 x + 2 y + 8 = 0$
centre $C_{2} = (4, - 1)$
and radius $r_{2} = \sqrt{4^{2} + 1^{2} - 8} = \sqrt{17 - 8} = 3$
Two circle intersect in two distinct points, then
$r_{1} - r_{2} < C_{1} C_{2} < r_{1} + r_{2}$
$r - 3 < \sqrt{(4 - 1)^{2} + (- 1 - 3)^{2}} < r + 3$
$r - 3 < \sqrt{9 + 16} < r + 3$
$r - 3 < 5 < r + 3$
$r < 8 and 2 < r$
$2 < r < 8$

AP EAMCET-20.04.2019

Conic Section

119992 The locus of the point of intersection of the tangents at the extremities of a chord of the circle $x^{2} + y^{2} = a^{2}$ which touches the circle $x^{2} + y^{2} - 2 a x = 0$ passes through the point

1

(a / 2, 0)

2

(0, a / 2)

3

(a, 0)

4

(0, 0)

Explanation:

A
Let the point of intersection of tangents be $(h, k)$
Then the chord of contact
$xh + yk = a^{2}$
The chord touched the circle
$x^{2} + y^{2} - 2 ax = 0$
centre $(a, 0)$ and Radius $= a$
Thus perpendicular from the centre on line is equal to radius of circle. Thus we get
$\sqrt{h^{2} + k^{2}} = \pm a$
$ah - a^{2} = a \sqrt{h^{2} + k^{2}}$
$h - a = \sqrt{h^{2} + k^{2}}$
Squaring both side we get
$(h - a)^{2} = h^{2} + k^{2}$
$h^{2} + a^{2} - 2 a h = h^{2} + k^{2}$
$k^{2} + 2 a h = a^{2}$
Locus of the point $y^{2} + 2 a x = a^{2}$
Hence, the curve passes through the point $(\frac{a}{2}, 0)$

AMU-2011

Conic Section

119993 The angle between circles $x^{2} + y^{2} + 2 x + 4 y + 1$ $= 0$ and $x^{2} + y^{2} - 2 x + 6 y - 3 = 0$ is:

1

\cos^{- 1} (\frac{3}{\sqrt{13}})

2

\cos^{- 1} (\frac{3}{\sqrt{31}})

3

\cos^{- 1} (\sqrt{\frac{3}{31}})

4

2 \cos^{- 1} (\frac{3}{\sqrt{13}})

Explanation:

A : Given,
$C_{1} : x^{2} + y^{2} + 2 x + 4 y + 1 = 0$
$C_{2} : x^{2} + y^{2} - 2 x + 6 y - 3 = 0$
$∴ C_{1} = (- 1, - 2) and r_{1} = \sqrt{1 + 4 - 1} = 2$
$C_{2} = (1, - 3) and r_{2} = \sqrt{1 + 9 + 3} = \sqrt{13}$
$∴ d = C_{1} C_{2} = \sqrt{(- 1 - 1)^{2} + (- 2 + 3)^{2}} = \sqrt{4 + 1} = \sqrt{5}$
$∴ \cos θ = | \frac{d^{2} - r_{1}^{2} - r_{2}^{2}}{2 r_{1} r_{2}} | = | \frac{5 - 4 - 13}{2 \times 2 \times \sqrt{13}} | = | \frac{- 3}{\sqrt{13}} |$
$∴ θ = \cos^{- 1} (\frac{3}{\sqrt{13}})$

APEAPCET-20.08.2021