QBManager

Conic Section

119994 The length of the common chord of the circles $x^{2} + y^{2} + 3 x + 5 y + 4 = 0$ and $x^{2} + y^{2} + 5 x + 3 y + 4$ $= 0$ is units.

1 3

2 2

3 6

4 4

Explanation:

D Given,
$C_{1} : x^{2} + y^{2} + 3 x + 5 y + 4 = 0$
$C_{2} : x^{2} + y^{2} + 5 x + 3 y + 4 = 0$
On solving these two circles, we get -
$x = y$
Substituting $x = y$ in equation (i), we get -
$2 x^{2} + 8 x + 4 = 0$
$x^{2} + 4 x + 2 = 0$
$(x + 2)^{2} = 2$
$x + 2 = \pm \sqrt{2}, x_{1, 2} = - 2 \pm \sqrt{2}$
$x = y = y_{1, 2} = - 2 \pm \sqrt{2}$
Length $= \sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}} = \sqrt{2} (x_{1} - x_{2})$
$= \sqrt{2} [(- 2 + \sqrt{2}) - (- 2 - \sqrt{2})] = 4$

AP EAPCET-25.08.2021

Conic Section

119995 Find the equation of the circle which passes through the point $(1, 2)$ and the points of intersection of the circles $x^{2} + y^{2} - 8 x - 6 y + 21 =$ 0 and $x^{2} + y^{2} - 2 x - 15 = 0$

1

x^{2} + y^{2} - 18 x - 12 y + 27 = 0

2

2 (x^{2} + y^{2}) - 18 x - 12 y + 27 = 0

3

3 (x^{2} + y^{2}) - 18 x - 12 y + 27 = 0

4

4 (x^{2} + y^{2}) - 18 x - 12 y + 27 = 0

Explanation:

C
$x^{2} + y^{2} - 8 x - 6 y + 21 + λ (x^{2} + y^{2} - 2 x - 15) = 0$
$(1)^{2} + (2)^{2} - 8 - 12 + 21 + λ ((1)^{2} + (2)^{2} - 2 \times 1 - 15) = 0$
$1 + 4 - 8 - 12 + 21 + λ (1 + 4 - 2 - 15) = 0$
$1 + 1 + 4 + λ (3 - 15) = 0$
$- 12 λ = 0$
$λ = 1 / 2$
Now,
$x^{2} + y^{2} - 8 x - 6 y + 21 + 1 / 2 (x^{2} + y^{2} - 2 x - 15) = 0$
$or 3 (x^{2} + y^{2}) - 18 x - 12 y + 27 = 0$

AP EAPCET-25.08.2021

Conic Section

119996 The perpendicular distance from the point $(1, 2)$ to common chord of the circles $x^{2} + y^{2} - 2 x$ $+ 4 y - 4 = 0$ and $x^{2} + y^{2} + 4 x - 6 y - 3 = 0$ is units.

1

\frac{13}{\sqrt{123}}

2

\frac{13}{\sqrt{136}}

3

\frac{13}{\sqrt{63}}

4

\frac{13}{\sqrt{132}}

Explanation:

B Given,
Ans: b
Exp: (B)Given,
$S_{1} \equiv x^{2} + y^{2} - 2 x + 4 y - 4 = 0$
$S_{2} \equiv x^{2} + y^{2} + 4 x - 6 y - 3 = 0$
$S_{1} - S_{2} = 0$
$\Rightarrow x^{2} + y^{2} - 2 x + 4 y - 4 - x^{2} - y^{2} - 4 x + 6 y + 3 = 0$
$\Rightarrow - 6 + 10 y - 1 = 0$
At, $(1, 2)$
$d = | \frac{{ax}_{1} + {by}_{1} + c}{\sqrt{a^{2} + b^{2}}} | = | \frac{- 6 + 20 - 1}{\sqrt{36 + 100}} | = \frac{+ 13}{\sqrt{136}}$

AP EAPCET-24.08.2021

Conic Section

119997 The radical centre of the circles $x^{2} + y^{2} + 3 x +$ $2 y + 1 = 0, x^{2} + y^{2} - x + 6 y + 5 = 0$ and $x^{2} + y^{2} +$ $5 x - 8 y + 15 = 0$ is

1

(3, 2)

2

(- 3, - 2)

3

(2, 3)

4

(- 2, - 3)

Explanation:

A Given,
$S_{1} : x^{2} + y^{2} + 3 x + 2 y + 1 = 0$
$S_{2} : x^{2} + y^{2} - x + 6 y + 5 = 0$
$S_{3} : x^{2} + y^{2} + 5 x - 8 y + 15 = 0$
Let, $(x, y)$ be the radical centre, then $S_{1} - S_{2} = 0$
$S_{2} - S_{3} = 0$
$4 x - 4 y - 4 = 0$ and $- 6 x + 14 y - 10 = 0$
$x = y + 1$ and $3 x - 7 y + 5 = 0$
$3 (y + 1) - 7 y + 5 = 0$
$3 y + 3 - 7 y + 5 = 0$
$- 4 y + 8 = 0$
$∴ y = 2$ and $x = 2 + 1 = 3$
Radical centre is $(3, 2)$ .

AP EAMCET-06.07.2022

Conic Section

119994 The length of the common chord of the circles $x^{2} + y^{2} + 3 x + 5 y + 4 = 0$ and $x^{2} + y^{2} + 5 x + 3 y + 4$ $= 0$ is units.

1 3

2 2

3 6

4 4

Explanation:

D Given,
$C_{1} : x^{2} + y^{2} + 3 x + 5 y + 4 = 0$
$C_{2} : x^{2} + y^{2} + 5 x + 3 y + 4 = 0$
On solving these two circles, we get -
$x = y$
Substituting $x = y$ in equation (i), we get -
$2 x^{2} + 8 x + 4 = 0$
$x^{2} + 4 x + 2 = 0$
$(x + 2)^{2} = 2$
$x + 2 = \pm \sqrt{2}, x_{1, 2} = - 2 \pm \sqrt{2}$
$x = y = y_{1, 2} = - 2 \pm \sqrt{2}$
Length $= \sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}} = \sqrt{2} (x_{1} - x_{2})$
$= \sqrt{2} [(- 2 + \sqrt{2}) - (- 2 - \sqrt{2})] = 4$

AP EAPCET-25.08.2021

Conic Section

119995 Find the equation of the circle which passes through the point $(1, 2)$ and the points of intersection of the circles $x^{2} + y^{2} - 8 x - 6 y + 21 =$ 0 and $x^{2} + y^{2} - 2 x - 15 = 0$

1

x^{2} + y^{2} - 18 x - 12 y + 27 = 0

2

2 (x^{2} + y^{2}) - 18 x - 12 y + 27 = 0

3

3 (x^{2} + y^{2}) - 18 x - 12 y + 27 = 0

4

4 (x^{2} + y^{2}) - 18 x - 12 y + 27 = 0

Explanation:

C
$x^{2} + y^{2} - 8 x - 6 y + 21 + λ (x^{2} + y^{2} - 2 x - 15) = 0$
$(1)^{2} + (2)^{2} - 8 - 12 + 21 + λ ((1)^{2} + (2)^{2} - 2 \times 1 - 15) = 0$
$1 + 4 - 8 - 12 + 21 + λ (1 + 4 - 2 - 15) = 0$
$1 + 1 + 4 + λ (3 - 15) = 0$
$- 12 λ = 0$
$λ = 1 / 2$
Now,
$x^{2} + y^{2} - 8 x - 6 y + 21 + 1 / 2 (x^{2} + y^{2} - 2 x - 15) = 0$
$or 3 (x^{2} + y^{2}) - 18 x - 12 y + 27 = 0$

AP EAPCET-25.08.2021

Conic Section

119996 The perpendicular distance from the point $(1, 2)$ to common chord of the circles $x^{2} + y^{2} - 2 x$ $+ 4 y - 4 = 0$ and $x^{2} + y^{2} + 4 x - 6 y - 3 = 0$ is units.

1

\frac{13}{\sqrt{123}}

2

\frac{13}{\sqrt{136}}

3

\frac{13}{\sqrt{63}}

4

\frac{13}{\sqrt{132}}

Explanation:

B Given,
Ans: b
Exp: (B)Given,
$S_{1} \equiv x^{2} + y^{2} - 2 x + 4 y - 4 = 0$
$S_{2} \equiv x^{2} + y^{2} + 4 x - 6 y - 3 = 0$
$S_{1} - S_{2} = 0$
$\Rightarrow x^{2} + y^{2} - 2 x + 4 y - 4 - x^{2} - y^{2} - 4 x + 6 y + 3 = 0$
$\Rightarrow - 6 + 10 y - 1 = 0$
At, $(1, 2)$
$d = | \frac{{ax}_{1} + {by}_{1} + c}{\sqrt{a^{2} + b^{2}}} | = | \frac{- 6 + 20 - 1}{\sqrt{36 + 100}} | = \frac{+ 13}{\sqrt{136}}$

AP EAPCET-24.08.2021

Conic Section

119997 The radical centre of the circles $x^{2} + y^{2} + 3 x +$ $2 y + 1 = 0, x^{2} + y^{2} - x + 6 y + 5 = 0$ and $x^{2} + y^{2} +$ $5 x - 8 y + 15 = 0$ is

1

(3, 2)

2

(- 3, - 2)

3

(2, 3)

4

(- 2, - 3)

Explanation:

A Given,
$S_{1} : x^{2} + y^{2} + 3 x + 2 y + 1 = 0$
$S_{2} : x^{2} + y^{2} - x + 6 y + 5 = 0$
$S_{3} : x^{2} + y^{2} + 5 x - 8 y + 15 = 0$
Let, $(x, y)$ be the radical centre, then $S_{1} - S_{2} = 0$
$S_{2} - S_{3} = 0$
$4 x - 4 y - 4 = 0$ and $- 6 x + 14 y - 10 = 0$
$x = y + 1$ and $3 x - 7 y + 5 = 0$
$3 (y + 1) - 7 y + 5 = 0$
$3 y + 3 - 7 y + 5 = 0$
$- 4 y + 8 = 0$
$∴ y = 2$ and $x = 2 + 1 = 3$
Radical centre is $(3, 2)$ .

AP EAMCET-06.07.2022

Conic Section

119994 The length of the common chord of the circles $x^{2} + y^{2} + 3 x + 5 y + 4 = 0$ and $x^{2} + y^{2} + 5 x + 3 y + 4$ $= 0$ is units.

1 3

2 2

3 6

4 4

Explanation:

D Given,
$C_{1} : x^{2} + y^{2} + 3 x + 5 y + 4 = 0$
$C_{2} : x^{2} + y^{2} + 5 x + 3 y + 4 = 0$
On solving these two circles, we get -
$x = y$
Substituting $x = y$ in equation (i), we get -
$2 x^{2} + 8 x + 4 = 0$
$x^{2} + 4 x + 2 = 0$
$(x + 2)^{2} = 2$
$x + 2 = \pm \sqrt{2}, x_{1, 2} = - 2 \pm \sqrt{2}$
$x = y = y_{1, 2} = - 2 \pm \sqrt{2}$
Length $= \sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}} = \sqrt{2} (x_{1} - x_{2})$
$= \sqrt{2} [(- 2 + \sqrt{2}) - (- 2 - \sqrt{2})] = 4$

AP EAPCET-25.08.2021

Conic Section

119995 Find the equation of the circle which passes through the point $(1, 2)$ and the points of intersection of the circles $x^{2} + y^{2} - 8 x - 6 y + 21 =$ 0 and $x^{2} + y^{2} - 2 x - 15 = 0$

1

x^{2} + y^{2} - 18 x - 12 y + 27 = 0

2

2 (x^{2} + y^{2}) - 18 x - 12 y + 27 = 0

3

3 (x^{2} + y^{2}) - 18 x - 12 y + 27 = 0

4

4 (x^{2} + y^{2}) - 18 x - 12 y + 27 = 0

Explanation:

C
$x^{2} + y^{2} - 8 x - 6 y + 21 + λ (x^{2} + y^{2} - 2 x - 15) = 0$
$(1)^{2} + (2)^{2} - 8 - 12 + 21 + λ ((1)^{2} + (2)^{2} - 2 \times 1 - 15) = 0$
$1 + 4 - 8 - 12 + 21 + λ (1 + 4 - 2 - 15) = 0$
$1 + 1 + 4 + λ (3 - 15) = 0$
$- 12 λ = 0$
$λ = 1 / 2$
Now,
$x^{2} + y^{2} - 8 x - 6 y + 21 + 1 / 2 (x^{2} + y^{2} - 2 x - 15) = 0$
$or 3 (x^{2} + y^{2}) - 18 x - 12 y + 27 = 0$

AP EAPCET-25.08.2021

Conic Section

119996 The perpendicular distance from the point $(1, 2)$ to common chord of the circles $x^{2} + y^{2} - 2 x$ $+ 4 y - 4 = 0$ and $x^{2} + y^{2} + 4 x - 6 y - 3 = 0$ is units.

1

\frac{13}{\sqrt{123}}

2

\frac{13}{\sqrt{136}}

3

\frac{13}{\sqrt{63}}

4

\frac{13}{\sqrt{132}}

Explanation:

B Given,
Ans: b
Exp: (B)Given,
$S_{1} \equiv x^{2} + y^{2} - 2 x + 4 y - 4 = 0$
$S_{2} \equiv x^{2} + y^{2} + 4 x - 6 y - 3 = 0$
$S_{1} - S_{2} = 0$
$\Rightarrow x^{2} + y^{2} - 2 x + 4 y - 4 - x^{2} - y^{2} - 4 x + 6 y + 3 = 0$
$\Rightarrow - 6 + 10 y - 1 = 0$
At, $(1, 2)$
$d = | \frac{{ax}_{1} + {by}_{1} + c}{\sqrt{a^{2} + b^{2}}} | = | \frac{- 6 + 20 - 1}{\sqrt{36 + 100}} | = \frac{+ 13}{\sqrt{136}}$

AP EAPCET-24.08.2021

Conic Section

119997 The radical centre of the circles $x^{2} + y^{2} + 3 x +$ $2 y + 1 = 0, x^{2} + y^{2} - x + 6 y + 5 = 0$ and $x^{2} + y^{2} +$ $5 x - 8 y + 15 = 0$ is

1

(3, 2)

2

(- 3, - 2)

3

(2, 3)

4

(- 2, - 3)

Explanation:

A Given,
$S_{1} : x^{2} + y^{2} + 3 x + 2 y + 1 = 0$
$S_{2} : x^{2} + y^{2} - x + 6 y + 5 = 0$
$S_{3} : x^{2} + y^{2} + 5 x - 8 y + 15 = 0$
Let, $(x, y)$ be the radical centre, then $S_{1} - S_{2} = 0$
$S_{2} - S_{3} = 0$
$4 x - 4 y - 4 = 0$ and $- 6 x + 14 y - 10 = 0$
$x = y + 1$ and $3 x - 7 y + 5 = 0$
$3 (y + 1) - 7 y + 5 = 0$
$3 y + 3 - 7 y + 5 = 0$
$- 4 y + 8 = 0$
$∴ y = 2$ and $x = 2 + 1 = 3$
Radical centre is $(3, 2)$ .

AP EAMCET-06.07.2022

Conic Section

119994 The length of the common chord of the circles $x^{2} + y^{2} + 3 x + 5 y + 4 = 0$ and $x^{2} + y^{2} + 5 x + 3 y + 4$ $= 0$ is units.

1 3

2 2

3 6

4 4

Explanation:

D Given,
$C_{1} : x^{2} + y^{2} + 3 x + 5 y + 4 = 0$
$C_{2} : x^{2} + y^{2} + 5 x + 3 y + 4 = 0$
On solving these two circles, we get -
$x = y$
Substituting $x = y$ in equation (i), we get -
$2 x^{2} + 8 x + 4 = 0$
$x^{2} + 4 x + 2 = 0$
$(x + 2)^{2} = 2$
$x + 2 = \pm \sqrt{2}, x_{1, 2} = - 2 \pm \sqrt{2}$
$x = y = y_{1, 2} = - 2 \pm \sqrt{2}$
Length $= \sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}} = \sqrt{2} (x_{1} - x_{2})$
$= \sqrt{2} [(- 2 + \sqrt{2}) - (- 2 - \sqrt{2})] = 4$

AP EAPCET-25.08.2021

Conic Section

119995 Find the equation of the circle which passes through the point $(1, 2)$ and the points of intersection of the circles $x^{2} + y^{2} - 8 x - 6 y + 21 =$ 0 and $x^{2} + y^{2} - 2 x - 15 = 0$

1

x^{2} + y^{2} - 18 x - 12 y + 27 = 0

2

2 (x^{2} + y^{2}) - 18 x - 12 y + 27 = 0

3

3 (x^{2} + y^{2}) - 18 x - 12 y + 27 = 0

4

4 (x^{2} + y^{2}) - 18 x - 12 y + 27 = 0

Explanation:

C
$x^{2} + y^{2} - 8 x - 6 y + 21 + λ (x^{2} + y^{2} - 2 x - 15) = 0$
$(1)^{2} + (2)^{2} - 8 - 12 + 21 + λ ((1)^{2} + (2)^{2} - 2 \times 1 - 15) = 0$
$1 + 4 - 8 - 12 + 21 + λ (1 + 4 - 2 - 15) = 0$
$1 + 1 + 4 + λ (3 - 15) = 0$
$- 12 λ = 0$
$λ = 1 / 2$
Now,
$x^{2} + y^{2} - 8 x - 6 y + 21 + 1 / 2 (x^{2} + y^{2} - 2 x - 15) = 0$
$or 3 (x^{2} + y^{2}) - 18 x - 12 y + 27 = 0$

AP EAPCET-25.08.2021

Conic Section

119996 The perpendicular distance from the point $(1, 2)$ to common chord of the circles $x^{2} + y^{2} - 2 x$ $+ 4 y - 4 = 0$ and $x^{2} + y^{2} + 4 x - 6 y - 3 = 0$ is units.

1

\frac{13}{\sqrt{123}}

2

\frac{13}{\sqrt{136}}

3

\frac{13}{\sqrt{63}}

4

\frac{13}{\sqrt{132}}

Explanation:

B Given,
Ans: b
Exp: (B)Given,
$S_{1} \equiv x^{2} + y^{2} - 2 x + 4 y - 4 = 0$
$S_{2} \equiv x^{2} + y^{2} + 4 x - 6 y - 3 = 0$
$S_{1} - S_{2} = 0$
$\Rightarrow x^{2} + y^{2} - 2 x + 4 y - 4 - x^{2} - y^{2} - 4 x + 6 y + 3 = 0$
$\Rightarrow - 6 + 10 y - 1 = 0$
At, $(1, 2)$
$d = | \frac{{ax}_{1} + {by}_{1} + c}{\sqrt{a^{2} + b^{2}}} | = | \frac{- 6 + 20 - 1}{\sqrt{36 + 100}} | = \frac{+ 13}{\sqrt{136}}$

AP EAPCET-24.08.2021

Conic Section

119997 The radical centre of the circles $x^{2} + y^{2} + 3 x +$ $2 y + 1 = 0, x^{2} + y^{2} - x + 6 y + 5 = 0$ and $x^{2} + y^{2} +$ $5 x - 8 y + 15 = 0$ is

1

(3, 2)

2

(- 3, - 2)

3

(2, 3)

4

(- 2, - 3)

Explanation:

A Given,
$S_{1} : x^{2} + y^{2} + 3 x + 2 y + 1 = 0$
$S_{2} : x^{2} + y^{2} - x + 6 y + 5 = 0$
$S_{3} : x^{2} + y^{2} + 5 x - 8 y + 15 = 0$
Let, $(x, y)$ be the radical centre, then $S_{1} - S_{2} = 0$
$S_{2} - S_{3} = 0$
$4 x - 4 y - 4 = 0$ and $- 6 x + 14 y - 10 = 0$
$x = y + 1$ and $3 x - 7 y + 5 = 0$
$3 (y + 1) - 7 y + 5 = 0$
$3 y + 3 - 7 y + 5 = 0$
$- 4 y + 8 = 0$
$∴ y = 2$ and $x = 2 + 1 = 3$
Radical centre is $(3, 2)$ .

AP EAMCET-06.07.2022

Question Bank Series

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