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(B) The equations of the circles are $S_1: x^2+y^2+2x+4y-3=0$ and $S_2: x^2+y^2+2kx-2y-1=0$.Comparing with $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g

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$4$

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(B) The equations of the circles are $S_1: x^2+y^2+2x+4y-3=0$ and $S_2: x^2+y^2+2kx-2y-1=0$.Comparing with $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$,we get:$g_1=1, f_1=2, c_1=-3$ and $g_2=k, f_2=-1, c_2=-1$.The centers are $C_1(-1, -2)$ and $C_2(-k, 1)$,and radii are $r_1=\sqrt{1^2+2^2-(-3)}=\sqrt{8}=2\sqrt{2}$ and $r_2=\sqrt{k^2+(-1)^2-(-1)}=\sqrt{k^2+2}$.The distance between centers $d^2 = (-k+1)^2 + (1+2)^2 = (k-1)^2 + 9 = k^2-2k+1+9 = k^2-2k+10$.The angle $	heta$ between the circles satisfies $\cos 	heta = \frac{d^2-r_1^2-r_2^2}{2r_1r_2}$.Given $\cos 	heta = \frac{1}{2\sqrt{3}}$,we have $\frac{k^2-2k+10-8-(k^2+2)}{2(2\sqrt{2})\sqrt{k^2+2}} = \frac{1}{2\sqrt{3}}$.$\frac{-2k}{4\sqrt{2}\sqrt{k^2+2}} = \frac{1}{2\sqrt{3}} \implies \frac{-k}{\sqrt{2}\sqrt{k^2+2}} = \frac{1}{\sqrt{3}}$.Squaring both sides: $\frac{k^2}{2(k^2+2)} = \frac{1}{3} \implies 3k^2 = 2k^2+4 \implies k^2=4$.

If the cosine of the angle between the two circles $x^2+y^2+2x+4y-3=0$ and $x^2+y^2+2kx-2y-1=0$ is $\frac{1}{2\sqrt{3}}$,then $k^2=$

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