If from any point on the circle x^2+y^2+2gx+2 | vedclass

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(C) Let the two circles be $C_1$ and $C_2$. $C_1: x^2+y^2+2gx+2fy+c=0$ with center $O(-g, -f)$ and radius $r_1 = \sqrt{g^2+f^2-c}$. $C_2: x^2+y^2+2gx+

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$2 \alpha$

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(C) Let the two circles be $C_1$ and $C_2$. $C_1: x^2+y^2+2gx+2fy+c=0$ with center $O(-g, -f)$ and radius $r_1 = \sqrt{g^2+f^2-c}$. $C_2: x^2+y^2+2gx+2fy+c \sin^2 \alpha + (g^2+f^2) \cos^2 \alpha = 0$. Rewriting $C_2$ as $(x+g)^2 + (y+f)^2 = g^2+f^2 - c \sin^2 \alpha - (g^2+f^2) \cos^2 \alpha$. $r_2^2 = g^2(1-\cos^2 \alpha) + f^2(1-\cos^2 \alpha) - c \sin^2 \alpha = (g^2+f^2-c) \sin^2 \alpha$. Thus,$r_2 = r_1 \sin \alpha$. Let $P$ be a point on $C_1$,so $OP = r_1$. Let $PA$ and $PB$ be tangents to $C_2$ from $P$. In $	riangle OAP$,$\angle OAP = 90^\circ$. $\sin(\angle OPA) = \frac{OA}{OP} = \frac{r_2}{r_1} = \frac{r_1 \sin \alpha}{r_1} = \sin \alpha$. Therefore,$\angle OPA = \alpha$. The angle between the tangents is $\angle APB = 2 \angle OPA = 2 \alpha$.

If from any point on the circle $x^2+y^2+2gx+2fy+c=0$,tangents are drawn to the circle $x^2+y^2+2gx+2fy+c \sin^2 \alpha + (g^2+f^2) \cos^2 \alpha = 0$,where $0 < \alpha < \frac{\pi}{2}$,then the angle between those tangents is

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