If the circles x^2+y^2-4x+2fy+1=0 and x^2+y^2 | vedclass

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(C) Given circles are $S_1: x^2+y^2-4x+2fy+1=0$ and $S_2: x^2+y^2+2gx-4y-1=0$.Comparing with $x^2+y^2+2g_ix+2f_iy+c_i=0$,we get:$g_1=-2, f_1=f, c_1=1$

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$2g^2$

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(C) Given circles are $S_1: x^2+y^2-4x+2fy+1=0$ and $S_2: x^2+y^2+2gx-4y-1=0$.Comparing with $x^2+y^2+2g_ix+2f_iy+c_i=0$,we get:$g_1=-2, f_1=f, c_1=1$$g_2=g, f_2=-2, c_2=-1$Since the circles cut orthogonally,the condition is $2(g_1g_2+f_1f_2)=c_1+c_2$.$2((-2)(g) + (f)(-2)) = 1 + (-1)$$2(-2g-2f) = 0 \implies g+f=0 \implies f=-g$.The radii squared are $r_1^2 = g_1^2+f_1^2-c_1 = (-2)^2+f^2-1 = 3+f^2$ and $r_2^2 = g_2^2+f_2^2-c_2 = g^2+(-2)^2-(-1) = g^2+5$.$r_1^2+r_2^2 = 3+f^2+g^2+5 = 8+f^2+g^2$.Since $f=-g$,$f^2=g^2$,so $r_1^2+r_2^2 = 8+2g^2$.Therefore,$r_1^2+r_2^2-8 = 2g^2$.

If the circles $x^2+y^2-4x+2fy+1=0$ and $x^2+y^2+2gx-4y-1=0$ cut orthogonally,then $r_1^2+r_2^2-8=$

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