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

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(C) Given circles are:$S_1: x^2+y^2+6x-2y+k=0$$S_2: x^2+y^2+2x-6y-15=0$If circle $S_1$ bisects the circumference of circle $S_2$,then the common chord

Vedclass · Accepted Answer

-$23$

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(C) Given circles are:$S_1: x^2+y^2+6x-2y+k=0$$S_2: x^2+y^2+2x-6y-15=0$If circle $S_1$ bisects the circumference of circle $S_2$,then the common chord of $S_1$ and $S_2$ must pass through the center of circle $S_2$.The equation of the common chord is given by $S_1 - S_2 = 0$:$(x^2+y^2+6x-2y+k) - (x^2+y^2+2x-6y-15) = 0$$4x+4y+k+15 = 0$ ... $(i)$The center of circle $S_2$ is $(-g, -f) = (-1, 3)$.Since the common chord $(i)$ passes through $(-1, 3)$,we substitute these coordinates into the equation:$4(-1) + 4(3) + k + 15 = 0$$-4 + 12 + k + 15 = 0$$8 + k + 15 = 0$$k + 23 = 0$$k = -23$Thus,the correct option is $C$.

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 $k=$

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