Question

The angle between the pair of tangents drawn from a point $P$ to the curve $x^2+y^2+4x-6y+9{\sin }^2\alpha +13{\cos }^2\alpha =0$ is $2\alpha$. 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

The correct option is D $x^2+y^2+4x-6y+9=0$

Center of the circle
$x^2+y^2+4x-6y+9{\sin }^2\alpha +13{\cos }^2\alpha =0$
is $C(-2,3)$ and its radius is
$\sqrt{{(-2)}^2+{\left(3\right)}^2-9{\sin }^2\alpha -13{\cos }^2\alpha }=\sqrt{13-13{\cos }^2\alpha -92{\sin }^2\alpha }=\sqrt{13{\sin }^2\alpha -9{\sin }^2\alpha }=\sqrt{4{\sin }^2\alpha }=2\sin \alpha$
Let $(h,k)$ be any point $P$ and $\angle APC=\alpha ,\angle PAC=\frac{\pi }{2}$
That is, triangle $APC$ is a right angle triangle
$\therefore \sin \alpha =\frac{AC}{PC}=\frac{2\sin \alpha }{\sqrt{{(h+2)}^2+{(k-3)}^2}}\Rightarrow {(h+2)}^2+{(k-3)}^2=4\Rightarrow h^2+4+4h+k^2+9-6k=4\Rightarrow h^2+y^2+4x-6y+9=0$
Thus, required equation of locus is
$x^2+y^2+4x-6y+9=0$