2012-I-17 – Solving Master

Ans: (a)

(x - 6)^{2} + (y - 10)^{2} = 10^{2}

(b)

(\frac{k - 4}{2}, \frac{k + 4}{2})

Since the $x$ -axis is a tangent to $C$ , then the radius is $10$ .
Hence, the equation of $C$ is

$\begin{array}{rcl} (x - 6)^{2} + (y - 10)^{2} & = & 10^{2} \\ x^{2} + y^{2} - 12 x - 20 y + 36 & = & 0 \end{array}$
Note that the equation of $L$ is $y = - x + k$ .
Substitute the equation of $L$ into the equation of $C$ , we have

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

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 $A B$

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

Therefore, the $y$ coordinates of the mid-point of $A B$

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

Therefore, the coordinates of the mid-point of $A B$ is $(\frac{k - 4}{2}, \frac{k + 4}{2})$ .