Question

The locus of the point of intersection of the tangents to the circle $x^2+y^2=a^2$ at the points whose parametric angle differ by $\frac{\pi }{3}$ is

• A. $2(x^2+y^2)=4a^2$
• B. $2(x^2+y^2)=a^2$
• C. $3(x^2+y^2)=4a^2$
• D. $3(x^2+y^2)=a^2$

Answer 1

The correct option is C $3(x^2+y^2)=4a^2$

Let the two points on the circle $x^2+y^2-a^2$ be $P(a\cos \theta ,b\sin \theta )$ and $\left[a\cos (\frac{\pi }{3}+\theta ),a\sin (\frac{\pi }{3}+\theta )\right].$

The equation of tangents at $P$ and $Q$ are

$x\cos \theta +y\sin \theta =a$ ...(1)

and, $x\cos (\frac{\pi }{3}+\theta )+y\sin (\frac{\pi }{3}+\theta )=a$

i.e., $\frac{1}{2}(x\cos \theta +y\sin \theta )-\frac{\sqrt{3}}{2}(x\sin \theta -y\cos \theta )=a$

or, $x\sin \theta -y\cos \theta =-\frac{a}{\sqrt{3}}$ ...(2) [Using (1)]

The locus of the point of intersection of the two tangents is obtained by eliminating (1) and (2), we get

${(x\cos \theta +y\sin \theta )}^2+{(x\sin \theta -y\cos \theta )}^2=a^2+\frac{a^2}{3}$

$\Rightarrow x^2+y^2=\frac{4a^2}{3}$ or $3(x^2+y^2)=4a^2,$