In the figure, $C$ is a variable circle of radius $1$. Let $G(u,v)$ be the center of the circle, parallelogram $OPQR$ is tangent to the circle at the point $R$, and circle cuts the $x$-axis at point $P$.
How to express coordinates of $P,Q,R$ in terms of $u,v$ and more importantly, how to find the locus of $Q$?
Edit: How to find the $\textbf{equation}$ of locus of $Q$?
Assuming that the circle always passes through the origin, here is a proposed solution.
Observe that the tangent $RQ$ is a horizontal tangent and horizontal tangents are possible only at points directly above and below centre on the circle. Therefore, coordinates of $R$ are $(u,1+v)$. Also, perpendicular from centre bisects a chord, therefore, $P$ is $(2u,0)$. Now, $OP=QR$, hence, $Q$ is $(3u,1+v)$.
Also, distance of centre from origin is $1$, so, $u^2+v^2=1$.
So, locus of $Q$ is an ellipse.
Hope it helps:)