2014-II-26

$\begin{array}{cl}
= & \left( – \dfrac{-8}{2}, – \dfrac{k}{2} \right) \\
= & (4, \dfrac{-k}{2})
\end{array}$

留意該直徑通過圓心及點 $(6,-5)$，且其斜率為 $-4$。由此，可得

$\begin{array}{rcl}
\dfrac{\frac{-k}{2} – (-5)}{4-6} & = & -4 \\
\dfrac{-k}{2} + 5 & = & 8 \\
\dfrac{-k}{2} & = & 3 \\
k & = & -6
\end{array}$