2012-I-17

Hence, the equation of $C$ is

$\begin{array}{rcl}
(x-6)^2+(y-10)^2 & = & 10^2 \\
x^2+y^2-12x-20y+36 & = & 0
\end{array}$

Substitute the equation of $L$ into the equation of $C$, we have

$\begin{array}{rcl}
x^2 +(-x+k)^2 -12x & & \\
-20(-x+k)+36 & = & 0 \\
2x^2+(8-2k)x+k^2 & & \\
-20k+36 & = & 0
\end{array}$

Note that the roots of the above equation are the $x$ coordinates of $A$ and $B$. Therefore the $x$ coordinates of the mid-point of $AB$

$\begin{array}{cl}
= & \dfrac{1}{2}\times\mbox{the sum of the roots} \\
= & \dfrac{1}{2} \times \left(-\dfrac{8-2k}{2}\right) \\
= & \dfrac{k-4}{2}
\end{array}$

Therefore, the $y$ coordinates of the mid-point of $AB$

$\begin{array}{cl}
= & -\dfrac{k-4}{2} +k \\
= & \dfrac{k+4}{2}
\end{array}$

Therefore, the coordinates of the mid-point of $AB$ is $\left( \dfrac{k-4}{2}, \dfrac{k+4}{2}\right)$.