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

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(D) Given,the equations of the circles are:$S: x^2+y^2+6x-2y+k=0$$S': x^2+y^2+2x-6y-15=0$The circle $S$ bisects the circumference of circle $S'$,which

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$-23$

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(D) Given,the equations of the circles are:$S: x^2+y^2+6x-2y+k=0$$S': x^2+y^2+2x-6y-15=0$The circle $S$ bisects the circumference of circle $S'$,which means the common chord of the two circles passes through the center of circle $S'$.The equation of the common chord is given by $S - S' = 0$:$(x^2+y^2+6x-2y+k) - (x^2+y^2+2x-6y-15) = 0$$4x + 4y + k + 15 = 0$  $(i)$The center of circle $S'$ is $(-g, -f)$,where $2g=2$ and $2f=-6$,so the center is $(-1, 3)$.Since the common chord passes through the center $(-1, 3)$,we substitute these coordinates into equation $(i)$:$4(-1) + 4(3) + k + 15 = 0$$-4 + 12 + k + 15 = 0$$8 + k + 15 = 0$$k + 23 = 0$$k = -23$Thus,option $D$ is correct.

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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