If the circle $x^{2} + y^{2} + 6 x - 2 y + k = 0$ bisects the circumference of the circle $x^{2} + y^{2} + 2 x - 6 y - 15 = 0$ , then k _______

-23

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-21

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Solution

The correct option is A
-23

Equation of the common chord of the given circles is

$4 x + 4 y + k + 15 = 0$ . . . .(1)

Since $x^{2} + y^{2} + 6 x - 2 y + k = 0$ bisects the circumference of $x^{2} + y^{2} + 2 x - 6 y = 15$

$∴$ (1) is the diameter of $x^{2} + y^{2} + 2 x - 6 y = 15$

$\Rightarrow 4 (- 1) + 4 (3) + k + 15 = 0$ $\Rightarrow k = - 23$

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