If the circle S equiv x^2+y^2+2gx+4y+1=0 bise | vedclass

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(B) The condition for circle $S_1 \equiv x^2+y^2+2g_1x+2f_1y+c_1=0$ to bisect the circumference of circle $S_2 \equiv x^2+y^2+2g_2x+2f_2y+c_2=0$ is th

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$\sqrt{12}$

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(B) The condition for circle $S_1 \equiv x^2+y^2+2g_1x+2f_1y+c_1=0$ to bisect the circumference of circle $S_2 \equiv x^2+y^2+2g_2x+2f_2y+c_2=0$ is that the common chord of the two circles must pass through the center of the circle being bisected. The common chord equation is $S_1 - S_2 = 0$. Given $S_1 \equiv x^2+y^2+2gx+4y+1=0$ and $S_2 \equiv x^2+y^2-2x-3=0$. The common chord is $(x^2+y^2+2gx+4y+1) - (x^2+y^2-2x-3) = 0$,which simplifies to $(2g+2)x + 4y + 4 = 0$. The center of the circle $S_2$ is $(1, 0)$. Since the common chord passes through $(1, 0)$,we substitute $x=1$ and $y=0$ into the equation: $(2g+2)(1) + 4(0) + 4 = 0$. $2g + 2 + 4 = 0 \implies 2g = -6 \implies g = -3$. The circle $S$ is $x^2+y^2-6x+4y+1=0$. The radius $r$ of the circle $x^2+y^2+2gx+2fy+c=0$ is $\sqrt{g^2+f^2-c}$. Here $g=-3, f=2, c=1$. $r = \sqrt{(-3)^2 + (2)^2 - 1} = \sqrt{9 + 4 - 1} = \sqrt{12}$.

If the circle $S \equiv x^2+y^2+2gx+4y+1=0$ bisects the circumference of the circle $x^2+y^2-2x-3=0$,then the radius of circle $S=0$ is

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