Ans: D
Denote $A$ the centre of the required circle and $N$ the mid-point of $PQ$. Therefore, $\angle ANP = 90^\circ$. Since $P$ and $Q$ are lying on the $x$-axis, the distance between $A$ and $PQ$ is $2$. Hence $AN = 2$. By applying the Pythagoras Theorem to $\Delta APN$, we have
$\begin{array}{rcl}
AP^2 & = & AN^2 + PN^2 \\
AP & = & \sqrt{2^2 + 3^2} \\
& = & \sqrt{13}
\end{array}$
Therefore, the radius of the required circle is $\sqrt{13}$. Hence the equation of the required circle is
$\begin{array}{rcl}
(x-(-5))^2 + (y-2)^2 & = & (\sqrt{13})^2 \\
x^2 + 10x +25 + y^2 -4y +4 – 13 & = & 0 \\
x ^ 2 +y^ 2 +10x – 4y +16 & = & 0
\end{array}$
