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 find $p + q$ .

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Solution

Given,

$x^{2} + y^{2} - 2 x + 8 y - q = 0$ ...(1)

$x^{2} + y^{2} + 4 x + 12 y + p = 0$ ...(2)

Let, $S_{1} \equiv x^{2} + y^{2} - 2 x + 8 y - q = 0 \to C (1, - 4)$

and $S_{2} \equiv x^{2} + y^{2} + 4 x + 12 y + p = 0$

equation of common chord of circles

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

$∴ x^{2} + y^{2} + 4 x + 12 y + p - (x^{2} + y^{2} - 2 x + 8 y - q) = 0$

$6 x + 4 y + p + q = 0$

$C_{1} = (1, - 4)$

$6 (1) + 4 (- 4) + p + q = 0$

$∴ p + q = 10$

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