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

Answer: (D)

We have equation of circle
$x^{2}+y^{2}+4 x-6 y+9 \sin ^{2} \alpha+13 \cos ^{2} \alpha=0$
Here, $C=(-2,3)$
Radius $=\sqrt{(-2)^{2}+(3)^{2}-\left(9 \sin ^{2} \alpha+13 \cos ^{2} \alpha\right)}$ $=\sqrt{4+9-9 \sin ^{2} \alpha-13 \cos ^{2} \alpha} $
$=\sqrt{13-13\left(1-\sin ^{2} \alpha\right)-9 \sin ^{2} \alpha} $
$=\sqrt{13 \sin ^{2} \alpha-9 \sin ^{2} \alpha} $
$=\sqrt{4 \sin ^{2} \alpha}=2 \sin \alpha $

Here, $\sin \alpha=\frac{A C}{P C} $
$\Rightarrow P C \sin \alpha=A C $
$\Rightarrow P C^{2} \sin ^{2} \alpha=A C^{2}=(2 \sin \alpha)^{2} $
$\Rightarrow {\left[(h+2)^{2}+(k-3)^{2}\right] \sin ^{2} \alpha=4 \sin ^{2} \alpha} $
$\Rightarrow (h+2)^{2}+(k-3)^{2}=4$
$\Rightarrow h^{2}+4+4 h+k^{2}+9-6 k=4$
$\Rightarrow h^{2}+k^{2}+4 h-6 k+9=0$
Hence, locus of a point is
$x^{2}+y^{2}+4 x-6 y+9=0$