Ans: C
$\left\{ \begin{array}{ll}
x^2 +y^2 -8y-14=0 & \ldots \unicode{x2460} \\
2x-y-6=0 & \ldots \unicode{x2461}
\end{array} \right.$
$\left\{ \begin{array}{ll}
x^2 +y^2 -8y-14=0 & \ldots \unicode{x2460} \\
2x-y-6=0 & \ldots \unicode{x2461}
\end{array} \right.$
From $\unicode{x2461}$, we have
$\begin{array}{rcl}
2x-y-6 & = & 0 \\
x & = & \dfrac{y+6}{2} \ \ldots \unicode{x2462}
\end{array}$
Sub. $\unicode{x2462}$ into $\unicode{x2460}$, we have
$\begin{array}{rcl}
\left(\dfrac{y+6}{2} \right)^2 +y^2 -8y -14 & = & 0 \\
y^2 +12y +36 +4y^2 -32y -56 & = & 0 \\
y^2 -4y -4 & = & 0 \\
\end{array}$
Therefore, the $y$-coordinate of the mid-point of $PQ$
$\begin{array}{cl}
= & \dfrac{1}{2} \times \left(- \dfrac{-4}{1}\right) \\
= & 2
\end{array}$
