Question

Let $P$ and $Q$ be points on the circle $x=a\cos \theta ,y=a\sin \theta$, such that the value of their parameters $\left(\theta \right)$ differs by $\frac{\pi }{6}$. The locus of the point of intersection of tangents to the circle at $P$ and $Q$ is

• A. $x^2+y^2=(8-4\sqrt{3})a^2$
• B. $x^2+y^2=(8-4\sqrt{2})a^2$
• C. $x^2+y^2=(6-4\sqrt{3})a^2$
• D. $x^2+y^2=(6-4\sqrt{2})a^2$

Answer 1

The correct option is A $x^2+y^2=(8-4\sqrt{3})a^2$


$\cos {15}^o=\frac{a}{\sqrt{h^2+k^2}}$
$\Rightarrow h^2+k^2=\frac{a^2}{\frac{1+\cos {30}^{\circ }}{2}}$

$\Rightarrow h^2+k^2=\frac{2a^2}{1+\frac{\sqrt{3}}{2}}$
$\Rightarrow h^2+k^2=\frac{4a^2}{2+\sqrt{3}}=4a^2(2-\sqrt{3})$
$\Rightarrow h^2+k^2=(8-4\sqrt{3})a^2$
$\Rightarrow x^2+y^2=(8-4\sqrt{3})a^2$