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.

$\because \tan \alpha =\frac{AC}{PA}$
$\tan \alpha =\frac{\sqrt{4+9-9{\sin }^2\alpha -13{\cos }^2\alpha }}{\sqrt{x_1^2+y_1^2+4x_1-6y_1+9{\sin }^2\alpha +13{\cos }^2\alpha }}$
$=\frac{\sqrt{13{\sin }^2\alpha -9{\sin }^2\alpha }}{\sqrt{x_1^2+y_1^2+4x_1-6y_1+9+4{\cos }^2\alpha }}$
$=\sqrt{\frac{4{\sin }^2\alpha }{x_1^2+y_1^2+4x_1-6y_1+9+4{\cos }^2\alpha }}$
$\Rightarrow \frac{{\sin }^2\alpha }{{\cos }^2\alpha }=\frac{4{\sin }^2\alpha }{x_1^2+y_1^2+4x_1-6y_1+9+4{\cos }^2\alpha }$
$\Rightarrow x_1^2+y_1^2+4x_1-6y_1+9+4{\cos }^2\alpha =4{\cos }^2\alpha$
$\Rightarrow x_1^2+y_1^2+4x_1-6y_1+9=0$
On taking locus of point $P\left(x_1,y_1\right)$, we get
$x^2+y^2+4x-6y+9=0$