Ans: B
$\left\{ \begin{array}{ll}
x^2+y^2+2x-2y-7=0 & \ldots \unicode{x2460}\\
3x-4y+k=0 & \ldots \unicode{x2461}
\end{array}\right.$
$\left\{ \begin{array}{ll}
x^2+y^2+2x-2y-7=0 & \ldots \unicode{x2460}\\
3x-4y+k=0 & \ldots \unicode{x2461}
\end{array}\right.$
From $\unicode{x2461}$, we have
$\begin{array}{rcl}
3x-4y+k & = & 0 \\
4y & = & 3x+k \ldots \unicode{x2462}
\end{array}$
Hence by $\unicode{x2462}$, we have
$\begin{array}{rcl}
x^2 + y^2 +2x-2y-7 & = & 0 \\
16x^2 + 16y^2 + 32x -32y -112 & = & 0 \\
16x^2 +(3x+k)^2 + 32x – 8(3x+k) – 112 & = & 0 \\
25x^2 +(8+6k)x+(k^2-8k-112) & = & 0
\end{array}$
For the circle and the straight line intersect,
$\begin{array}{rcl}
\Delta & \ge & 0 \\
(8+6k)^2 – 4(25)(k^2-8k-112) & \ge & 0 \\
– 64k^2 +896k+11264 & \ge & 0 \\
k^2 -14k -176 & \le & 0 \\
(k+8)(k-22) & \le & 0
\end{array}$
Therefore, $-8\le k \le 22$.
