Ans: C
$\left\{ \begin{array}{ll}
x-y=k & \ldots \unicode{x2460} \\
x^2 + y^2 +2x -4y -1=0 & \ldots \unicode{x2461}
\end{array} \right.$
$\left\{ \begin{array}{ll}
x-y=k & \ldots \unicode{x2460} \\
x^2 + y^2 +2x -4y -1=0 & \ldots \unicode{x2461}
\end{array} \right.$
From $\unicode{x2460}$, we have
$\begin{array}{rcl}
x-y & = & k \\
y & = & x-k \ \ldots \unicode{x2462}
\end{array}$
Sub. $\unicode{x2462}$ into $\unicode{x2461}$, we have
$\begin{array}{rcl}
x^2 +(x-k)^2 +2x -4(x-k)-1 & = & 0 \\
x^2 +x^2 -2kx +k^2 +2x -4x +4k -1 & = & 0 \\
2x^2 +(-2k -2)x + k^2 +4k -1 & = & 0 \ \ldots \unicode{x2463}
\end{array}$
Let $x_1$ and $x_2$ be the $x$-coordinates of $A$ and $B$ respectively. Therefore, $x_1$ and $x_2$ are the roots of equation $\unicode{x2463}$.
The $x$-coordinate of the mid-point of $AB$
$\begin{array}{cl}
= & \dfrac{x_1+x_2}{2} \\
= & \dfrac{1}{2} \times \left(- \dfrac{-2k-2}{2} \right) \\
= & \dfrac{1+k}{2}
\end{array}$
