留意 $(1,0)$ 在 $hx+ky=6$ 之上，可得
$\begin{array}{rcl}
h(1)+k(0) & = & 6 \\
h & = & 6
\end{array}$
由此，可得
$\left\{ \begin{array}{ll}
6x+ky=6 & \ldots \unicode{x2460} \\
x^2+y^2-8x-4y-18=0 & \ldots \unicode{x2461}
\end{array}\right.$
從 $\unicode{x2460}$，可得
$\begin{array}{rcl}
6x+ky & = & 6 \\
x & = & \dfrac{6-ky}{6} \ \ldots \unicode{x2462}
\end{array}$
把 $\unicode{x2462}$ 代入 $\unicode{x2461}$，可得
$\begin{array}{rcl}
\left(\dfrac{6-ky}{6}\right)^2+y^2-8\left(\dfrac{6-ky}{6}\right)-4y-18 & = & 0 \\
\dfrac{36-12ky+k^2y^2}{36}+y^2+\dfrac{-48+8ky}{6}-4y-18 & = & 0 \\
k^2y^2-12ky+36+36y^2-288+48ky-144y-648 & = & 0 \\
(36+k^2)y^2+(36k-144)y-900 & = & 0
\end{array}$
由於 $MN$ 中點的坐標為 $(1,0)$，可得
$\begin{array}{rcl}
\dfrac{\text{兩根之和s}}{2} & = & 0 \\
\dfrac{-(36k-144)}{2(36+k^2)} & = & 0 \\
-36k+144 & = & 0 \\
k & = & 4
\end{array}$
