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

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

$x^2+y^2+4x-6y+9{\sin }^2\alpha +13{\cos }^2\alpha =0$
$C(-2,3),r=\sqrt{4+9-9{\sin }^2\alpha -13{\cos }^2\alpha }=\sqrt{13{\sin }^2\alpha -9{\sin }^2\alpha }=2\sin \alpha$
$\sin \alpha =\frac{AC}{PC}$
${PC}^2{\sin }^2\alpha =4{\sin }^2\alpha$
${(h+2)}^2+{(k-3)}^2=4$
$\therefore$ locus of $P$ is $x^2+y^2+4x-6y+9=0$