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

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(D) The condition for a circle $S_1 = x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ to bisect the circumference of another circle $S_2 = x^2 + y^2 + 2g_2x + 2f

Vedclass · Accepted Answer

$-23$

Vedclass · Answer

(D) The condition for a circle $S_1 = x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ to bisect the circumference of another circle $S_2 = x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ is that the common chord must pass through the center of the second circle.The common chord is given by $S_1 - S_2 = 0$,which is $(6-2)x + (-2+6)y + (k+15) = 0$,or $4x + 4y + k + 15 = 0$.The center of the second circle $x^2 + y^2 + 2x - 6y - 15 = 0$ is $(-g_2, -f_2) = (-1, 3)$.Substituting $(-1, 3)$ into the equation of the common chord: $4(-1) + 4(3) + k + 15 = 0$.$-4 + 12 + k + 15 = 0$.$8 + k + 15 = 0$.$k + 23 = 0 \Rightarrow k = -23$.

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 $k =$

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