Question

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