The angle between the pair of tangents drawn from a point P to the circle and its locus

The angle between the 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 $$2\alpha$$ . Then the equation of the locus of the point 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- $$x^2+y^2+4x−6y+9=0$$

My approach is as follow

$${x^2} + {y^2} + 4x - 6y + 9{\sin ^2}\alpha + 13{\cos ^2}\alpha = 0 \Rightarrow {\left( {x + 2} \right)^2} + {\left( {y - 3} \right)^2} = 13{\sin ^2}\alpha - 9{\sin ^2}\alpha = {\left( {2\left| {\sin \alpha } \right|} \right)^2} = {R^2}$$

$${X^2} + {Y^2} = {R^2}$$

$$Y = mX \pm \sqrt {{{\mathop{\rm R}\nolimits} ^2} + {R^2}{m^2}}$$ is the general equation of the tangent

$$\Rightarrow Y - mX = \pm R\sqrt {1 + {m^2}} \Rightarrow {Y^2} + {m^2}{X^2} - 2mYX = {R^2} + {R^2}{m^2}$$

$$\Rightarrow {m^2}\left( {{X^2} - {R^2}} \right) - 2mYX + {Y^2} - {R^2} = 0$$

$$\tan 2\alpha = \frac{{{m_1} + {m_2}}}{{1 - {m_1}{m_2}}} = \frac{{\frac{{2YX}}{{\left( {{X^2} - {R^2}} \right)}}}}{{1 - \frac{{\left( {{Y^2} - {R^2}} \right)}}{{\left( {{X^2} - {R^2}} \right)}}}} = \frac{{2YX}}{{{X^2} - {Y^2}}}$$

$$2\sin \alpha = R \Rightarrow \sin \alpha = \frac{R}{2} \Rightarrow \tan \alpha = \frac{R}{{\sqrt {4 - {R^2}} }}$$

$$\frac{{2\tan \alpha }}{{1 - {{\tan }^2}\alpha }} = \frac{{2YX}}{{{X^2} - {Y^2}}} = \frac{{2\left( {y - 3} \right)\left( {x + 2} \right)}}{{{{\left( {x + 2} \right)}^2} - {{\left( {y - 3} \right)}^2}}}$$

not able to proceed from here

On drawing a diagram, we see that $$P$$ is at a constant distance from center $$O=(-2,3)$$. You can now finish.