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:

  1. 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 $



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