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If from any point on the circle $x^2+y^2+2 g x+2 f y+c=0$, tangents are drawn to the circle $x^2+y^2+2 g x+2 f y+c \sin ^2 \alpha$ $+\left(g^2+f^2\right) \cos ^2 \alpha=0,\left(0 < \alpha < \frac{\pi}{2}\right)$, then the angle between those tangents is
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Verified Answer
The correct answer is:
$\frac{\pi}{3}$
Given equations of circles are
$$
c_1 \equiv x^2+y^2+2 g x+2 f y+c=0
$$
Centre $O(-g,-f)$
$$
\begin{aligned}
& \text { radius } r_1=\sqrt{g^2+f^2-c} \\
& \begin{aligned}
c_2 \equiv x^2+y^2+ & 2 g x+2 f y \\
& +c \sin ^2 \alpha+\left(g^2+f^2\right) \cos ^2 \alpha
\end{aligned}
\end{aligned}
$$
Centre $O(-g,-f)$
$$
\text { radius } r_2=\sqrt{g^2+f^2-c \sin ^2 \alpha-g^2 \cos ^2 \alpha-f^2 \cos ^2 \alpha}
$$
$$
r_2=\sqrt{g^2\left(1-\cos ^2 \alpha\right)+f^2\left(1-\cos ^2 \alpha\right)-c \sin ^2 \alpha}
$$
$$
=\sqrt{g^2+f^2-c} \cdot \sin \alpha
$$
as $0 < \alpha < \pi / 2 \Rightarrow 0 < \sin \alpha < 1 \Rightarrow r_2 < r_1$
Circles $c_1$ and $c_2$ are concentric circles
Now, $r_2=\sqrt{g^2+f^2-c} \sin \alpha$
$r_2=r_1 \sin \alpha$
$$
c_1 \equiv x^2+y^2+2 g x+2 f y+c=0
$$
Centre $O(-g,-f)$
$$
\begin{aligned}
& \text { radius } r_1=\sqrt{g^2+f^2-c} \\
& \begin{aligned}
c_2 \equiv x^2+y^2+ & 2 g x+2 f y \\
& +c \sin ^2 \alpha+\left(g^2+f^2\right) \cos ^2 \alpha
\end{aligned}
\end{aligned}
$$
Centre $O(-g,-f)$
$$
\text { radius } r_2=\sqrt{g^2+f^2-c \sin ^2 \alpha-g^2 \cos ^2 \alpha-f^2 \cos ^2 \alpha}
$$
$$
r_2=\sqrt{g^2\left(1-\cos ^2 \alpha\right)+f^2\left(1-\cos ^2 \alpha\right)-c \sin ^2 \alpha}
$$
$$
=\sqrt{g^2+f^2-c} \cdot \sin \alpha
$$
as $0 < \alpha < \pi / 2 \Rightarrow 0 < \sin \alpha < 1 \Rightarrow r_2 < r_1$
Circles $c_1$ and $c_2$ are concentric circles
Now, $r_2=\sqrt{g^2+f^2-c} \sin \alpha$
$r_2=r_1 \sin \alpha$
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