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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=

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$- 21$

$23$

$- 23$

$21$

answer is D.

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Detailed Solution

$s - s^{1} = 0$

$\Rightarrow 4 x + 4 y + k + 15 = 0$

$c_{2} (- 1, 3) \Rightarrow - 4 + 12 + k + 15 = 0$

$\Rightarrow k = - 23$

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