答案：B
$\left\{ \begin{array}{ll}
x^2+y^2+2x-4y-13=0 & \ldots \unicode{x2460} \\
x-y+k=0 & \ldots \unicode{x2461}
\end{array}\right.$
$\left\{ \begin{array}{ll}
x^2+y^2+2x-4y-13=0 & \ldots \unicode{x2460} \\
x-y+k=0 & \ldots \unicode{x2461}
\end{array}\right.$
從 $\unicode{x2461}$，可得
$\begin{array}{rcl}
x-y+k & = & 0 \\
y & = & x+k~\ldots \unicode{x2462}
\end{array}$
把 $\unicode{x2462}$ 代入 $\unicode{x2460}$，可得
$\begin{array}{rcl}
x^2+(x+k)^2+2x-4(x+k)-13 & = & 0 \\
2x^2+ (-2+2k)x +(k^2-4k-13) & = & 0
\end{array}$
由於該直線與圓交於兩點，可得
$\begin{array}{rcl}
\Delta & > & 0 \\
(-2+2k)^2-4(2)(k^2-4k-13) & > & 0 \\
k^2 -6k -27 & < & 0 \\
(k-9)(k+3) & < & 0 \\
\end{array}$
所以，$-3 < k < 9$。
