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

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(D) Given that,the circle $S_1 \equiv x^2+y^2+6x-2y+k=0$ and $S_2 \equiv x^2+y^2+2x-6y-15=0$. Since $S_1$ bisects the circumference of $S_2$,the commo

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-$23$

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(D) Given that,the circle $S_1 \equiv x^2+y^2+6x-2y+k=0$ and $S_2 \equiv x^2+y^2+2x-6y-15=0$. 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 $S_1 - S_2 = 0$. $(x^2+y^2+6x-2y+k) - (x^2+y^2+2x-6y-15) = 0$. $4x + 4y + k + 15 = 0$. Since this chord is the diameter of $S_2$,it must pass through the center of $S_2$. The center of $S_2$ is $(-g, -f) = (-1, 3)$. 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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