Circle 2 Question 12
13. The angle between a pair of tangents drawn from a point $P$ 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 $P$ is
(1996, 1M)
(a) $x^{2}+y^{2}+4 x-6 y+4=0$
(b) $x^{2}+y^{2}+4 x-6 y-9=0$
(c) $x^{2}+y^{2}+4 x-6 y-4=0$
(d) $x^{2}+y^{2}+4 x-6 y+9=0$
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Answer:
Correct Answer: 13. (d)
Solution:
- Centre of the circle
$x^{2}+y^{2}+4 x-6 y+9 \sin ^{2} \alpha+13 \cos ^{2} \alpha=0$
is $C(-2,3)$ and its radius is
$ \begin{aligned} & \sqrt{(-2)^{2}+(3)^{2}-9 \sin ^{2} \alpha-13 \cos ^{2} \alpha} \\ & \quad=\sqrt{13-13 \cos ^{2} \alpha-9 \sin ^{2} \alpha} \\ & \quad=\sqrt{13 \sin ^{2} \alpha-9 \sin ^{2} \alpha}=\sqrt{4 \sin ^{2} \alpha}=2 \sin \alpha \end{aligned} $
Let $(h, k)$ be any point $P$ and
$ \angle A P C=\alpha, \angle P A C=\pi / 2 $
That is, triangle $A P C$ is a right angled triangle.
$ \begin{array}{lc} \therefore & \sin \alpha=\frac{A C}{P C}=\frac{2 \sin \alpha}{\sqrt{(h+2)^{2}+(k-3)^{2}}} \\ \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 \end{array} $
Thus, required equation of the locus is
$ x^{2}+y^{2}+4 x-6 y+9=0 $
