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Orthogonality of Two Circles

Match the fol...

Question

Match the following
$I$ : lf $x^{2} + y^{2} - 6 x - 8 y + 12 = 0, a) 2$
$x^{2} + y^{2} - 4 x + 6 y + k = 0$
cut orthogonally then $k =$
$I I$ lf $x^{2} + y^{2} - 2 x + 3 y + k = 0, b) - 10$
$x^{2} + y^{2} + 8 x - 6 y - 7 = 0$
cut orthogonally then $k =$
$I I I$ : lf $x^{2} + y^{2} + 2 x - 2 y + 4 = 0, c) - 24$
$x^{2} + y^{2} + 4 x - 2 y + k = 0$
cut orthogonally then $k =$

A

a, b, c

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B

b, c, a

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C

c, b, a

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D

a, c, b

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Solution

The correct option is A $c, b, a$
$I . x^{2} + y^{2} - 6 x - 8 y + 12 = 0 a n d x^{2} + y^{2} - 4 x + 6 y + k = 0$
$4 \equiv (3, 4) r a d i u s r_{1} = \sqrt{13}$
If this given circles cuts orthogonally then
$2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2}$
So, $20 (- 3) (- 2) + 2 (- 6) (3) = 12 + k$
$12 - 24 = 12 + k$
$k = - 24$
$I I . x^{2} + y^{2} - 2 x + 3 y + k = 0 a n d x 2 + y^{2} + 8 x - 6 y - 7 = 0$ cuts each othor othogonally than
$2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2}$
So, $2 x (- 1) (4) + 2 (^{1} /_{2}) (- 3) = k - 7$
$- 8 - 9 = k - 7$
$k = - 10$
$I I I x^{2} + y^{2} + 2 x - 2 y + 4 = 0 a n d x^{2} + y^{2} + 4 x - 2 y + 2 = 0$ are orthogonal to each other , so
$2 x (1) (2) + 2 x (- 1) (- 1) = 4 + 2$
$6 = 6$
True

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Similar questions

A

x^{2} + y^{2} - 2 x + 3 y + k = 0

x^{2} + y^{2} + 8 x - 6 y - 7 = 0

cut each other orthogonally, then

k = 10

R :

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

x^{2} + y^{2} + 2 g^{'} x + 2 f^{'} y + c^{'} = 0

cut each other orthogonally if

2 g g^{'} + 2 f f^{'} = c + c^{'}

Q. If the circle

x^{2} + y^{2} + 8 x - 4 y + c = 0

touches the circle

x^{2} + y^{2} + 2 x + 4 y - 11 = 0

externally and cuts the circle

x^{2} + y^{2} - 6 x + 8 y + k = 0

orthogonally then

k =

Q. For what value of

k

x^{2} + y^{2} + 5 x + 3 y + 7 = 0

x^{2} + y^{2} - 8 x + 6 y + k = 0

cut orthogonally.

Circles	Number of common tangents
I: $x^{2} + y^{2} = 4$ $x^{2} + y^{2} - 8 x + 12 = 0$	a) $0$
II: $x^{2} + y^{2} = 1$ $x^{2} + y^{2} - 2 x - 6 y + 6 = 0$	b) $1$
III: $x^{2} + y^{2} = 16$ $x^{2} + y^{2} - 8 x + 6 y - 56 = 0$	c) $4$
IV: $x^{2} + y^{2} - 2 x - 6 y + 9 = 0$ $x^{2} + y^{2} + 6 x - 2 y + 1 = 0$	d) $3$

Q. If the locus of centre of circle which cuts the circles

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

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

orthogonally is

a x + b y + c = 0

a + b + c

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