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

Answer 1

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

The equation of a given circle is
$x^2+y^2+4x-6y+9{\sin }^2\alpha +13{\cos }^2\alpha =0$
$\therefore$ Centre $A(-2,3)$

In $△ABP$,
$\sin \alpha =\frac{2\sin \alpha }{\sqrt{(h+2{)}^2+(k-3{)}^2}}$
$\Rightarrow (h+2{)}^2+(k-3{)}^2=4$
$\Rightarrow h^2+k^2+4h-6k+9=0$
Hence, the required locus of $p(h,k)$ is
$x^2+y^2+4x-6y+9=0$