Question

The angle between a pair of tangents drawn from a point $T$ to the circle $x^2+y^2+4x-6y+9{\sin }^2\theta +13{\cos }^2\theta =0$ is $2\sin \theta$. The equation of the locus of the point $T$ 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 $x^2+y^2+4x-6y+9=0$

$x^2+y^2+4x-6y+9{\sin }^2\theta +13{\cos }^2\theta =0$
centre $(-2,3)$
radius = $\sqrt{13-13{\cos }^2\theta -9{\sin }^2\theta }$
$=\left|2\sin \theta \right|$
In $\Delta$ CAT,
$CT\cos (90-\theta )=CA$
$CT\sin \theta =\left|2\sin \theta \right|$
$CT=2$
$\sqrt{(h+2{)}^2+(k-3{)}^2}=2$
$\Rightarrow h^2+k^2+4h-6k+9=0$
so, locus of $T(h,k)$ is
$\Rightarrow x^2+y^2+4x-6y+9=0$