If the angle between a pair of tangents drawn from a point $P$ to the circle $x^2+y^2+4 x-6 y+9 \sin ^2 \alpha+13 \cos ^2 \alpha=0$ is $2 \alpha$, then the equation of the locus of $P$ is

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Aicharya · Accepted Answer

According to given information, on drawing the figure.

$\begin{aligned} & \because \quad \tan \alpha=\frac{A C}{P A} \\ & \Rightarrow \tan \alpha=\frac{\sqrt{4+9-9 \sin ^2 \alpha-13 \cos ^2 \alpha}}{\sqrt{x_1^2+y_1^2+4 x_1-6 y_1+9 \sin ^2 \alpha+13 \cos ^2 \alpha}} \\ & =\frac{\sqrt{13 \sin ^2 \alpha-9 \sin ^2 \alpha}}{\sqrt{x_1^2+y_1^2+4 x_1-6 y_1+9+4 \cos ^2 \alpha}} \\ & =\sqrt{\frac{4 \sin ^2 \alpha}{x_1^2+y_1^2+4 x_1-6 y_1+9+4 \cos ^2 \alpha}} \\ & \Rightarrow \frac{\sin ^2 \alpha}{\cos ^2 \alpha}=\frac{4 \sin ^2 \alpha}{x_1^2+y_1^2+4 x_1-6 y_1+9+4 \cos ^2 \alpha} \\ & \Rightarrow x_1^2+y_1^2+4 x_1-6 y_1+9+4 \cos ^2 \alpha=4 \cos ^2 \alpha \\ & \Rightarrow x_1^2+y_1^2+4 x_1-6 y_1+9=0\end{aligned}$ On taking locus of point $P\left(x_1, y_1\right)$, we get $x^2+y^2+4 x-6 y+9=0$ Hence, option (4) is correct.

If the angle between a pair of tangents drawn from a point \(P\) to the circle \(x^2+y^2+4 x-6 y+9 \sin ^2…

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