We claim the desired circle is the circle whose center is the midpoint $O$ of $AB$ and that touches $BC$. We use complex numbers with $a=-1,b=1$. Then by intersection formula, $p=\frac{2cq+q-c}{c+q},l=\frac 12(b+p)=\frac{q(c+1)}{c+q}$. Now the reflection of $O$ over $QL$ is given by $\frac{\bar lq-l\bar q}{l-q}=\frac{-(c+1)}{q}$. The distance of this point to $O$ is equal to $|c+1|$ and thus fixed.

Let $M$ be midpoint of $AB$. we claim that our circle is a circle with center $M$ which touches $BC$. Note that we need to prove $M$ lies on external angle bisector of $\angle PLQ$. from Thales Theorem we have $ML || AQ$ so $\angle MLP = \angle APB = \angle LPQ$ so we need to prove $PL = QL$. Note that $PL = LB$ so we need to prove $\angle PQB = \angle 90$ which is true.