Question

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

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

Answer 1

Answer: (D)

We have equation of circle
$x^2+y^2+4x-6y+9{\sin }^2\alpha +13{\cos }^2\alpha =0$
Here, $C=(-2,3)$
Radius $=\sqrt{(-2{)}^2+(3{)}^2-\left(9{\sin }^2\alpha +13{\cos }^2\alpha \right)}$ $=\sqrt{4+9-9{\sin }^2\alpha -13{\cos }^2\alpha }$
$=\sqrt{13-13\left(1-{\sin }^2\alpha \right)-9{\sin }^2\alpha }$
$=\sqrt{13{\sin }^2\alpha -9{\sin }^2\alpha }$
$=\sqrt{4{\sin }^2\alpha }=2\sin \alpha$

Here, $\sin \alpha =\frac{AC}{PC}$
$\Rightarrow PC\sin \alpha =AC$
$\Rightarrow P{C}^2{\sin }^2\alpha =A{C}^2=(2\sin \alpha {)}^2$
$\Rightarrow \left[(h+2{)}^2+(k-3{)}^2\right]{\sin }^2\alpha =4{\sin }^2\alpha$
$\Rightarrow (h+2{)}^2+(k-3{)}^2=4$
$\Rightarrow h^2+4+4h+k^2+9-6k=4$
$\Rightarrow h^2+k^2+4h-6k+9=0$
Hence, locus of a point is
$x^2+y^2+4x-6y+9=0$