圆 $S_1$ 的圆心为 $C$，其方程为 $x^2 + y^2 - 8x - 16y + 55 = 0$。 另一个圆 $S_2$ 的圆心为 $D$，且经过 $C$、$A$ 和 $B$。

1. Group the $x$ and $y$ terms and prepare to complete the square

$x^2 - 8x + y^2 - 16y = -55$

2. Complete the square for both $x$ and $y$ to find the standard form of the circle equation

$(x - 4)^2 - 16 + (y - 8)^2 - 64 = -55$
$(x - 4)^2 + (y - 8)^2 = 16 + 64 - 55$
$(x - 4)^2 + (y - 8)^2 = 25$

3. Extract the centre coordinates and radius from the standard form

Centre $C = (4, 8)$
Radius $r = \sqrt{25} = 5$

1. Find the gradient of the radius $CA$

From part (a), the centre of $S_1$ is $C(4, 8)$.
Gradient of $CA = \frac{12 - 8}{7 - 4} = \frac{4}{3}$

2. Find the gradient of the tangent at $A$

Since the tangent is perpendicular to the radius at the point of contact,
Gradient of tangent $= -\frac{1}{\frac{4}{3}} = -\frac{3}{4}$

3. Find the equation of the tangent

Using the point $A(7, 12)$ and gradient $-\frac{3}{4}$:
$y - 12 = -\frac{3}{4}(x - 7)$
$4(y - 12) = -3(x - 7)$
$4y - 48 = -3x + 21$
$3x + 4y = 69$

1. Use the equation of the tangent from part (b) to find the $x$-intercept

From part (b), the equation of the tangent is $3x + 4y = 69$.
At the $x$-axis, $y = 0$.
$3x + 4(0) = 69$
$3x = 69$
$x = 23$

1. Identify the geometric relationship between points $C$, $A$, and $B$

From earlier parts, $C(4, 8)$ is the centre of $S_1$, $A(7, 12)$ is on $S_1$, and $AB$ is the tangent to $S_1$ at $A$. Therefore, the radius $CA$ is perpendicular to the tangent $AB$, meaning $\angle CAB = 90^\circ$.

2. Deduce that $CB$ is the diameter of $S_2$

Since $S_2$ passes through $C$, $A$, and $B$, and $\angle CAB = 90^\circ$, $CB$ must be the diameter of $S_2$ (angle in a semicircle is a right angle).

3. Find the coordinates of $D$, the centre of $S_2$

The centre $D$ is the midpoint of the diameter $CB$. Using $C(4, 8)$ and $B(23, 0)$:
$D = \left( \frac{4 + 23}{2}, \frac{8 + 0}{2} \right) = \left( \frac{27}{2}, 4 \right)$ (Shown)

4. Calculate the square of the radius of $S_2$

The radius $r_2$ is the distance $CD$:
$r_2^2 = \left( \frac{27}{2} - 4 \right)^2 + (4 - 8)^2$
$r_2^2 = \left( \frac{19}{2} \right)^2 + (-4)^2 = \frac{361}{4} + 16 = \frac{425}{4}$

5. Write the equation of $S_2$

Using the centre $D\left(\frac{27}{2}, 4\right)$ and $r_2^2 = \frac{425}{4}$:
$\left(x - \frac{27}{2}\right)^2 + (y - 4)^2 = \frac{425}{4}$

1. Calculate the distance between the centre of $S_1$ and point $D$

From part (a), the centre of $S_1$ is $C(4, 8)$ and its radius is $r = 5$.
Distance $CD = \sqrt{\left(\frac{27}{2} - 4\right)^2 + (4 - 8)^2}$
$CD = \sqrt{9.5^2 + (-4)^2}$
$CD = \sqrt{90.25 + 16} = \sqrt{106.25} \approx 10.3$

2. Compare the distance with the radius to determine the position of $D$

Since $CD = \sqrt{106.25} > 5$, the distance from the centre is greater than the radius.
Therefore, $D$ lies outside $S_1$.