Question

Tangents are drawn one to each of the concentric circles $x^2+y^2=4$ and $x^2+y^2=9$. They include an angle of ${60}^{\circ }$. Find the locus of the point of intersection of these tangents.

Answer 1

Let tangents at $A$ and $B$ meet at $P(h,k)$ and include an angle ${60}^{\circ }$.
If $\angle CPA=\theta$ then $\angle CPB=\theta +{60}^{\circ }$
$\therefore \sin \theta =\frac{2}{CP},\sin (\theta +{60}^{\circ })=\frac{3}{CP}$
$\therefore \frac{\sin (\theta +{60}^{\circ })}{\sin \theta }=\frac{3}{2}$
or $2(\frac{1}{2}\sin \theta +\frac{\sqrt{3}}{2}\cos \theta )=3\sin \theta$
or $2\sin \theta =\sqrt{3}\cos \theta$ or $\tan \theta =\frac{\sqrt{3}}{2}$
But $\tan \theta =\frac{CA}{PA}=\frac{2}{\sqrt{S^{\prime }}}=\frac{2}{\sqrt{h^2+k^2-4}}$
$\therefore \frac{\sqrt{3}}{2}=\frac{2}{\sqrt{h^2+k^2-4}}$. Square and generalize.
$3(x^2+y^2-4)=16$
or $3(x^2+y^2)=28$
It is again a concentric circle.