If the circle x^2+y^2+6x-2y+k=0 bisects the c | vedclass

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(D) Given the circles:$S_1 \equiv x^2+y^2+6x-2y+k=0$$S_2 \equiv x^2+y^2+2x-6y-15=0$Since $S_1$ bisects the circumference of $S_2$,the common chord of

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$-23$

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(D) Given the circles:$S_1 \equiv x^2+y^2+6x-2y+k=0$$S_2 \equiv x^2+y^2+2x-6y-15=0$Since $S_1$ bisects the circumference of $S_2$,the common chord of $S_1$ and $S_2$ must be the diameter of $S_2$.The equation of the common chord is given by $S_1 - S_2 = 0$:$(x^2+y^2+6x-2y+k) - (x^2+y^2+2x-6y-15) = 0$$4x + 4y + k + 15 = 0$The center of circle $S_2$ is $(-g, -f) = (-1, 3)$.Since the common chord is the diameter of $S_2$,it must pass through the center of $S_2$ $(-1, 3)$.Substituting $(-1, 3)$ into the chord equation:$4(-1) + 4(3) + k + 15 = 0$$-4 + 12 + k + 15 = 0$$8 + k + 15 = 0$$k + 23 = 0$$k = -23$

If the circle $x^2+y^2+6x-2y+k=0$ bisects the circumference of the circle $x^2+y^2+2x-6y-15=0$,then $k$ is equal to :

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