A circle S = x^2 + y^2 + 2gx + 2fy + 4 = 0 cu | vedclass

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(A) For two circles $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ to cut orthogonally,the condition is $2g_1g_2 + 2f

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$4$

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(A) For two circles $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ to cut orthogonally,the condition is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$. Given $S: x^2 + y^2 + 2gx + 2fy + 4 = 0$ and $C_1: x^2 + y^2 - 4x - 4y - 4 = 0$. Applying the condition: $2g(-2) + 2f(-2) = 4 - 4$ $\Rightarrow -4g - 4f = 0$ $\Rightarrow g + f = 0$. Let $r_1$ be the radius of $S$,so $r_1^2 = g^2 + f^2 - 4$. Given $S$ makes an angle of $60^{\circ}$ with $C_2: x^2 + y^2 + 4x + 4y + 4 = 0$. The centers are $O_1 = (-g, -f)$ and $O_2 = (-2, -2)$. The distance $d^2 = (-g + 2)^2 + (-f + 2)^2 = (g - 2)^2 + (f - 2)^2$. The radius of $C_2$ is $r_2 = \sqrt{2^2 + 2^2 - 4} = 2$. The angle $	heta$ between circles is given by $\cos 	heta = \frac{r_1^2 + r_2^2 - d^2}{2r_1r_2}$. $\cos 60^{\circ} = \frac{1}{2} = \frac{r_1^2 + 4 - ((g - 2)^2 + (f - 2)^2)}{2r_1(2)}$. $2r_1 = r_1^2 + 4 - (g^2 - 4g + 4 + f^2 - 4f + 4) = r_1^2 + 4 - (g^2 + f^2 - 4(g + f) + 8)$. Since $g + f = 0$ and $g^2 + f^2 = r_1^2 + 4$,we have $2r_1 = r_1^2 + 4 - (r_1^2 + 4 - 0 + 8) = r_1^2 + 4 - r_1^2 - 12 = -8$. Taking the magnitude,$2r_1 = 8$,so $r_1 = 4$.

$A$ circle $S = x^2 + y^2 + 2gx + 2fy + 4 = 0$ cuts the circle $x^2 + y^2 - 4x - 4y - 4 = 0$ orthogonally and makes an angle of $60^{\circ}$ with the circle $x^2 + y^2 + 4x + 4y + 4 = 0$. Then the radius of the circle $S = 0$ is

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