A circle S equiv x^2+y^2+2gx+2fy+6=0 cuts ano | vedclass

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(A) Two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ intersect orthogonally if $2(g_1g_2+f_1f_2) = c_1+c_2$. For the first pair

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(A) Two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ intersect orthogonally if $2(g_1g_2+f_1f_2) = c_1+c_2$. For the first pair: $2(g(-3) + f(-3)) = 6 - 6 = 0$,so $g+f=0$,which means $f=-g$. The circle $S$ is $x^2+y^2+2gx-2gy+6=0$. The radius $r_1 = \sqrt{g^2+(-g)^2-6} = \sqrt{2g^2-6}$. The second circle is $x^2+y^2+6x+6y+2=0$ with center $C_2(-3, -3)$ and radius $r_2 = \sqrt{3^2+3^2-2} = \sqrt{16} = 4$. The distance between centers $C_1(-g, g)$ and $C_2(-3, -3)$ is $d^2 = (-g+3)^2 + (g+3)^2 = g^2-6g+9 + g^2+6g+9 = 2g^2+18$. Using the cosine rule for the angle $	heta = 60^{\circ}$ between circles: $\cos 	heta = \frac{r_1^2+r_2^2-d^2}{2r_1r_2}$. $\cos 60^{\circ} = \frac{1}{2} = \frac{(2g^2-6) + 16 - (2g^2+18)}{2 \cdot r_1 \cdot 4} = \frac{-8}{8r_1} = -\frac{1}{r_1}$. This implies $r_1 = -2$,which is impossible as radius must be positive. Re-evaluating the angle: The angle between circles is the angle between their tangents at the point of intersection. If the angle is $60^{\circ}$,$\cos 60^{\circ} = \frac{1}{2}$. Given the calculation $\frac{1}{2} = \frac{2g^2-6+16-2g^2-18}{8r_1} = \frac{-8}{8r_1} = -\frac{1}{r_1}$,there might be a sign convention issue in the problem statement or the angle is $120^{\circ}$. Assuming the magnitude leads to $r_1 = 2$.

$A$ circle $S \equiv x^2+y^2+2gx+2fy+6=0$ cuts another circle $x^2+y^2-6x-6y-6=0$ orthogonally. If the angle between the circles $S=0$ and $x^2+y^2+6x+6y+2=0$ is $60^{\circ}$,then the radius of the circle $S=0$ is

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