直线 $y = 10$ 和 $3y + 4x = 32$ 分别在点 $(-2, 10)$ 和 $(2, 8)$ 处与圆 $C$ 相切。

1. Find the equation of the normal to the circle at $(-2, 10)$.

The tangent is $y = 10$, which is a horizontal line.
The normal at $(-2, 10)$ is perpendicular to the tangent, so it is a vertical line passing through $x = -2$.
Equation of the normal: $x = -2$.

2. Find the gradient of the second tangent and its normal.

The second tangent is $3y + 4x = 32$, which can be rewritten as $y = -\frac{4}{3}x + \frac{32}{3}$.
Gradient of the tangent, $m_1 = -\frac{4}{3}$.
Gradient of the normal, $m_2 = \frac{-1}{m_1} = \frac{3}{4}$.

3. Find the equation of the normal to the circle at $(2, 8)$.

Using the point $(2, 8)$ and gradient $m_2 = \frac{3}{4}$:
$y - 8 = \frac{3}{4}(x - 2)$
$y = \frac{3}{4}x - \frac{3}{2} + 8$
$y = \frac{3}{4}x + \frac{13}{2}$

4. Find the coordinates of the centre of the circle.

The centre of the circle is the intersection of the two normals.
Substitute $x = -2$ into $y = \frac{3}{4}x + \frac{13}{2}$:
$y = \frac{3}{4}(-2) + \frac{13}{2} = -\frac{3}{2} + \frac{13}{2} = \frac{10}{2} = 5$.
Centre of the circle is $(-2, 5)$.

5. Find the radius of the circle and write its equation.

The radius $r$ is the distance from the centre $(-2, 5)$ to the point $(-2, 10)$.
$r = \sqrt{(-2 - (-2))^2 + (10 - 5)^2} = \sqrt{0 + 5^2} = 5$.
$r^2 = 25$.
The equation of the circle is $(x - (-2))^2 + (y - 5)^2 = 25$, which simplifies to $(x + 2)^2 + (y - 5)^2 = 25$ (shown).

1. Identify the centre and radius of the circle from its equation.

From the equation $(x + 2)^2 + (y - 5)^2 = 25$, the centre of the circle is $(-2, 5)$ and the radius is $\sqrt{25} = 5$.

2. Compare the distance from the centre to the x-axis with the radius to conclude.

The distance from the centre $(-2, 5)$ to the x-axis ($y = 0$) is $5$ units.
Since this distance is equal to the radius of the circle, the x-axis is tangent to $C$.

1. Identify the centre and radius of the circle from part (a)

From $(x + 2)^2 + (y - 5)^2 = 25$, the centre is $(-2, 5)$ and the radius is $r = \sqrt{25} = 5$.

2. Determine the equations of the vertical tangents

Vertical tangents are of the form $x = c$. They occur at a horizontal distance of $r$ from the centre's x-coordinate.
$x = -2 - 5 = -7$
$x = -2 + 5 = 3$

1. Identify the radius of the circle and the length of half the chord

From part (a), the equation of the circle is $(x + 2)^2 + (y - 5)^2 = 25$.
The radius of the circle is $r = \sqrt{25} = 5$ units.
The perpendicular from the centre to a chord bisects the chord, so half the length of $PQ$ is $\frac{4}{2} = 2$ units.

2. Calculate the shortest distance using Pythagoras' theorem

Let $d$ be the shortest distance from the centre to the line $PQ$.
Using Pythagoras' theorem in the right-angled triangle formed by the radius, half the chord, and the perpendicular distance:
$d^2 + 2^2 = 5^2$
$d^2 + 4 = 25$
$d^2 = 21$
$d = \sqrt{21} \approx 4.58$ units (to 3 s.f.)