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 $2 \alpha$. The equation of the locus of the point ${ }^{'} P$ ' is

• A

  $x^{2}+y^{2}+4 x-6 y+4=0$
• B

  $x^{2}+y^{2}+4 x-6 y-9=0$
• C

  $x^{2}+y^{2}+4 x-6 y-4=0$
• Correct answer
  D

  $x^{2}+y^{2}+4 x-6 y+9=0$

Solution

The equation of a given circle is $$ x^{2}+y^{2}+4 x-6 y+9 \sin ^{2} \alpha+13 \cos ^{2} \alpha=0 $$ $\therefore$ Centre $A(-2,3)$ in $\Delta A B P$, $$ \sin \alpha=\frac{2 \sin \alpha}{\sqrt{(h+2)^{2}+(k-3)^{2}}} $$
$\Rightarrow \quad(h+2)^{2}+(k-3)^{2}=4$ $\Rightarrow \quad h^{2}+k^{2}+4 h-6 k+9=0$ Hence, the required locus of $P(h, k)$ is $$ x^{2}+y^{2}+4 x-6 y+9=0 $$

Hence, the correct answer is option: 'D'.