A problem related to locus in circles Two circles intersect at $A$ and $B$; $PQ$ is a straight line through $A$ meeting the circles at $P$ and $Q$. Find the locus of midpoint of $PQ$.
I know the locus is a circle. But I am unable to prove it.

Depends how you want to prove this, but going to polar coordinates with origin $A$ makes for a pretty straightforward proof.
"Polar coordinates"
For any point $x\in\mathbb R^2$ there exist $\rho\in\mathbb R$ and a unit vector $\mathbf e$ such that $x=A+\rho\mathbf e$. Note that I don't restrict $\rho$ to be positive so a point can be expressed by several different values of $(\rho,\mathbf e)$. It doesn't really matter since we can still express all points under that form.
Circle equation
Now consider some circle with center $c$ and radius $r$, such that $A$ belongs to the circle. The circle is the collection of points $x=A+\rho\mathbf e$ such that
$$
\rho=r\left\langle\mathbf e,\ \frac{c-A}{\left\|c-A\right\|}\right\rangle
=\left\langle\mathbf e,\ c-A\right\rangle
$$
Locus
Now let's say you have two circles with centers $c_1$ and $c_2$, and radii
$r_1$ and $r_2$. One line $PQ$ corresponds to a fixed direction $\mathbf e$. I assume that whenever possible, $P\neq A\neq Q$. This means that
$P$ and $Q$ are the points $x_1=A+\rho_1\mathbf e$ and $x_2=A+\rho_2\mathbf e$, where $\rho_i=\langle\mathbf e,\ c_i-A\rangle$.
The midpoint of $P$ and $Q$ is
$A+\rho\mathbf e$ where
$\rho=\frac{\rho_1+\rho_2}2$. It's easy to see there is $c$ such that
$\rho=\langle\mathbf e,\ c-A\rangle$, that suffices to prove your locus is the circle of center $c$ and radius $r=\|c-A\|$.
In detail:
\begin{align*}
\rho &=
\left\langle \mathbf e,\ \frac{c_1-A+c_2-A}{2} \right\rangle
=\left\langle \mathbf e,\ \frac{c_1+c_2}{2}-A \right\rangle
=\left\langle \mathbf e,\ c-A \right\rangle
\end{align*}
So just take $c$ as the midpoint between $c_1$ and $c_2$, your locus is the circle centered at $c$ that goes through $A$.
