Question

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

Answer: (D)

$C$ is $(-2,3)$
and $r_{2}^{2}=4+9-9 \sin ^{2} \alpha-13 \cos ^{2} \alpha$
$=4-4 \cos ^{2} \alpha=4 \sin ^{2} \alpha$
$\therefore \frac{ CP }{ CT }=\sin \alpha=\frac{2 \sin \alpha}{ CT }$
$\therefore CT =2=$ constant
Thus the point $T$ is at a constant distance $2$ from $C (-2,3)$.
Hence locus of $T$ is $( x +2)^{2}+( y -3)^{2}=2^{2}$
or $x^{2}+y^{2}-4 x+6 y+9=0$