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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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The correct answer is:
\(x^2+y^2+4 x-6 y+9=0\)
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.
\(\begin{aligned} & \because \quad \tan \alpha=\frac{A C}{P A} \\ & \Rightarrow \tan \alpha=\frac{\sqrt{4+9-9 \sin ^2 \alpha-13 \cos ^2 \alpha}}{\sqrt{x_1^2+y_1^2+4 x_1-6 y_1+9 \sin ^2 \alpha+13 \cos ^2 \alpha}} \\ & =\frac{\sqrt{13 \sin ^2 \alpha-9 \sin ^2 \alpha}}{\sqrt{x_1^2+y_1^2+4 x_1-6 y_1+9+4 \cos ^2 \alpha}} \\ & =\sqrt{\frac{4 \sin ^2 \alpha}{x_1^2+y_1^2+4 x_1-6 y_1+9+4 \cos ^2 \alpha}} \\ & \Rightarrow \frac{\sin ^2 \alpha}{\cos ^2 \alpha}=\frac{4 \sin ^2 \alpha}{x_1^2+y_1^2+4 x_1-6 y_1+9+4 \cos ^2 \alpha} \\ & \Rightarrow x_1^2+y_1^2+4 x_1-6 y_1+9+4 \cos ^2 \alpha=4 \cos ^2 \alpha \\ & \Rightarrow x_1^2+y_1^2+4 x_1-6 y_1+9=0\end{aligned}\)
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.
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