To find tangent at a point $Q$

Question

So Question is

The tangent at any point $P$ on the parabola $y^2=4ax$ intersects the $y$-axis at $Q$. Then tangent to the circumcircle of $\triangle PQS$ ($S$ is the focus) at $Q$ is:

(A) a line parallel to $x$-axis (B) $y$-axis
(C) a line parallel to $y$-axis (D) none of these

I assumed point $P$ as $(at^2,2at)$ then from there I got equation of tangent to parabola later I got point $Q$ as $(0,at)$. After having $P,Q,S$ points I am not understanding how to find tangent at $Q$.

Circumcenter I got is $(a(t^2+1)/2, at)$

Answer 1

Slope of the radius joining the circumcentre and $Q$ is $$m=\frac{at-at}{a(t^2+1)-0}=0$$The tangent at $Q$ is perpendicular to this radius, and hence has the slope $\infty$, i.e. the tangent is parallel to the $y$ axis. But the tangent passes through $Q$ which itself lies on $y$-axis, and thus the tangent is the $y$-axis.