The circumference of the circle x^2 + y^2 - 2 | vedclass

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(C) Let the two circles be $S_1: x^2 + y^2 - 2x + 8y - q = 0$ and $S_2: x^2 + y^2 + 4x + 12y + p = 0$.The common chord of the two circles is given by

Vedclass · Accepted Answer

$10$

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(C) Let the two circles be $S_1: x^2 + y^2 - 2x + 8y - q = 0$ and $S_2: x^2 + y^2 + 4x + 12y + p = 0$.The common chord of the two circles is given by $S_1 - S_2 = 0$.$(x^2 + y^2 - 2x + 8y - q) - (x^2 + y^2 + 4x + 12y + p) = 0$$-6x - 4y - q - p = 0$$6x + 4y + (p + q) = 0$Since the circumference of $S_1$ is bisected by $S_2$,the common chord must be the diameter of $S_1$.The center of $S_1$ is $(1, -4)$.Since the diameter passes through the center $(1, -4)$,we substitute these coordinates into the equation of the common chord:$6(1) + 4(-4) + (p + q) = 0$$6 - 16 + (p + q) = 0$$-10 + (p + q) = 0$$p + q = 10$

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

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