A circle with center at origin passes through three points $P$, $Q$ and $R$ with the line segment $PQ$ as its diameter along $x$-axis. A line passes through $P$ intersects the chord $QR$ at point $D$. The angle between the line segment $PD$ and $x$-axis is $\alpha$ and the distance between point $P$ and the midpoint of $DQ$ is $d$.
If $PD$ is the bisector of the angle RPQ then find the radius of the circle.
I took coordinates of $Q$ as $(a,0)$ and $P$ as $(-a,0)$. Equation of PD will be $y=\tan\alpha(x+a)$.
So we can take coordinate of $D$ as $(h,\tan\alpha(h+a))$ If mid point of $DQ$ is $S$, then we get coordinates of $S$ in terms of $h,a,\alpha$ but how would be eliminate $h$ as radius is $'a'$ units and $\alpha$ is mentioned in the question, so we somehow need to eliminate $h$ to find radius in terms of $d$ and $\alpha$. How should I proceed?