<s>[INST] 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
 [/INST] 
The choices should alert you that a simpler method is possible.
On drawing a diagram, we see that $P$ is at a constant distance from center $O=(-2,3)$. You can now finish.
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