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Particular Solution of a Differential Equation

The circumfer...

Question

The circumference of the circle $x^{2} + y^{2} - 2 x + 8 y - q = 0$ is bisected by the circle $x^{2} + y^{2} + 4 x + 12 y + p = 0$ , then $p + q$ is equal to

25

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100

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10

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48

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Solution

The correct option is B $10$
Common chord of the 2 circles is given by

$S_{1} - S_{2} = 0$

$⟹ 6 x + 4 y + (p + q) = 0$

Since $S_{2}$ bisects $S_{1}$ , center of $S_{1}$ lies on common chord.

$⟹ 6 (1) + 4 (- 4) + p + q = 0$

$p + q = 10$

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