I am not sure how to continue or if this is correct.
Let the centre of the circle be (h,k)
$(k-3)^2+(h+1)^2=10$ ... (1)
$(k-5)^2+(h-3)^2=10$...(2)
Solving (1) and (2), yields k+2h = 6 ....(3)
Equation of the perpendicular bisector of PQ: $y = -2x+6$
Substituting (h,k) yields $k = -2h + 6 $....(4)
I tried to solve (3) and (4), but it does not work. Help is needed. Thanks.
Here is a minimally algebraic solution that happens to rely on the specific geometry of the problem.
We know that the center $O$ will lie on the perpendicular bisector of $PQ$, and we know that $|OP|^2 = |OQ|^2 = 10$. The length of $PQ$ is $$|PQ| = \sqrt{(3-(-1))^2 + (5-3)^2} = \sqrt{20} = \sqrt{2}\sqrt{10}.$$ So it follows that $$|OP|^2 + |OQ|^2 = |PQ|^2,$$ hence there are two possible solutions for the center of the circle, say $O$ and $O'$, and $POQO'$ is a square. Since the midpoint of $PQ$ is $M = (1, 4)$, it is not difficult to see that a rotation of $P, Q$ about $M$ by $\pi/2$ will yield the points $O, O' \in \{(0,6), (2,2)\}$ in some order. These are the desired centers, and the corresponding equations of the circle follow in the obvious way.
