圆 $C_1$ 的最高点是 (2, 8)。在点 (6, 6) 处与 $C_1$ 相切的切线 $T$ 的方程为 $3y + 4x = 42$。

1. Determine the $x$-coordinate of the centre of the circle.

Since the highest point on the circle is $(2, 8)$, the $x$-coordinate of the centre must be $2$. Let the centre be $(2, b)$.

2. Find the gradient of the normal to the circle at $(6, 6)$.

The equation of the tangent is $3y + 4x = 42$, which can be rewritten as $y = -\frac{4}{3}x + 14$.
The gradient of the tangent is $-\frac{4}{3}$.
The gradient of the normal is $\frac{-1}{-\frac{4}{3}} = \frac{3}{4}$.

3. Find the $y$-coordinate of the centre.

The normal passes through the point of tangency $(6, 6)$ and the centre $(2, b)$.
$\frac{6 - b}{6 - 2} = \frac{3}{4}$
$\frac{6 - b}{4} = \frac{3}{4}$
$6 - b = 3 \implies b = 3$
The centre of the circle is $(2, 3)$.

4. Calculate the radius squared and write the equation of the circle.

The radius squared, $r^2$, is the squared distance between the centre $(2, 3)$ and the point $(6, 6)$.
$r^2 = (6 - 2)^2 + (6 - 3)^2 = 4^2 + 3^2 = 16 + 9 = 25$
The equation of the circle $C_1$ is $(x - 2)^2 + (y - 3)^2 = 25$.

1. Recall the center and radius of $C_1$ from part (a) and establish the geometric relationship for the reflection.

From part (a), the center of $C_1$ is $(2, 3)$ and its radius is $5$.
Since $C_2$ is the reflection of $C_1$ in the tangent $T$ at the point $(6, 6)$, the point of tangency $(6, 6)$ is the midpoint of the line segment joining the centers of $C_1$ and $C_2$.

2. Find the center of $C_2$ using the midpoint formula.

Let the center of $C_2$ be $(x_2, y_2)$.
Using the midpoint formula:
$\frac{2 + x_2}{2} = 6 \implies 2 + x_2 = 12 \implies x_2 = 10$
$\frac{3 + y_2}{2} = 6 \implies 3 + y_2 = 12 \implies y_2 = 9$
The center of $C_2$ is $(10, 9)$.

3. Write the equation of $C_2$.

The radius of $C_2$ is the same as $C_1$, which is $5$.
The equation of $C_2$ is:
$(x - 10)^2 + (y - 9)^2 = 5^2$
$(x - 10)^2 + (y - 9)^2 = 25$