Ans: B
$\left\{ \begin{array}{ll}
x^2 +y^2+2x-6y+k=0 & \ldots \unicode{x2460} \\
x + y + 4=0 & \ldots \unicode{x2461}
\end{array} \right.$
$\left\{ \begin{array}{ll}
x^2 +y^2+2x-6y+k=0 & \ldots \unicode{x2460} \\
x + y + 4=0 & \ldots \unicode{x2461}
\end{array} \right.$
From $\unicode{x2461}$, we have
$\begin{array}{rcl}
x + y + 4 & = & 0 \\
y & = & -x – 4 \ \ldots \unicode{x2462}
\end{array}$
Sub. $\unicode{x2462}$ into $\unicode{x2460}$, we have
$\begin{array}{rcl}
x^2 +(-x-4)^2 +2x -6(-x-4) + k & = & 0 \\
x^2 +x^2 +8x + 16 +2x +6x +24 +k & = & 0 \\
2x^2 + 16x + (40+k) & = & 0
\end{array}$
Since the circle and the straight line intersect at only one point, we have
$\begin{array}{rcl}
\Delta & = & 0 \\
16^2 – 4(2)(40+k) & = & 0 \\
256 -320 – 8k & = & 0 \\
8k & = & -64 \\
k & = & -8
\end{array}$
