Question

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

• A. $x^2 + y^2 + 4x -6y + 4 = 0$
• B. $x^2 + y^2 + 4x - 6y - 9 = 0$
• C. $x^2 + y^2 - 4x + 6y - 4 = 0$
• D. $x^2 + y^2 + 4x - 6y + 9 = 0$

Answer 1

Answer: (D)

According to given information, on drawing the figure.

$\because \tan \alpha=\frac{A C}{P A}$
$\tan \alpha=\frac{\sqrt{4+9-9 \sin ^{2} \alpha-13 \cos ^{2} \alpha}}{\sqrt{x_{1}^{2}+y_{1}^{2}+4 x_{1}-6 y_{1}+9 \sin ^{2} \alpha+13 \cos ^{2} \alpha}}$
$=\frac{\sqrt{13 \sin ^{2} \alpha-9 \sin ^{2} \alpha}}{\sqrt{x_{1}^{2}+y_{1}^{2}+4 x_{1}-6 y_{1}+9+4 \cos ^{2} \alpha}}$
$=\sqrt{\frac{4 \sin ^{2} \alpha}{x_{1}^{2}+y_{1}^{2}+4 x_{1}-6 y_{1}+9+4 \cos ^{2} \alpha}}$
$\Rightarrow \frac{\sin ^{2} \alpha}{\cos ^{2} \alpha}=\frac{4 \sin ^{2} \alpha}{x_{1}^{2}+y_{1}^{2}+4 x_{1}-6 y_{1}+9+4 \cos ^{2} \alpha}$
$\Rightarrow x_{1}^{2}+y_{1}^{2}+4 x_{1}-6 y_{1}+9+4 \cos ^{2} \alpha=4 \cos ^{2} \alpha$
$\Rightarrow x_{1}^{2}+y_{1}^{2}+4 x_{1}-6 y_{1}+9=0$
On taking locus of point $P\left(x_{1}, y_{1}\right)$, we get
$x^{2}+y^{2}+4 x-6 y+9=0$