圆 $C$ 的方程为 $x^2 + y^2 + 6x - 4y = 12$。

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

$x^2 + 6x + y^2 - 4y = 12$

2. Complete the square for both $x$ and $y$

$(x + 3)^2 - 3^2 + (y - 2)^2 - (-2)^2 = 12$
$(x + 3)^2 - 9 + (y - 2)^2 - 4 = 12$

3. Rearrange into the standard circle equation form $(x-a)^2 + (y-b)^2 = r^2$

$(x + 3)^2 + (y - 2)^2 = 12 + 9 + 4$
$(x + 3)^2 + (y - 2)^2 = 25$

4. Identify the centre and radius from the standard form

Centre is $(-3, 2)$.
Radius $r = \sqrt{25} = 5$.

1. State the centre of the circle $C$ from part (i)

From part (i), the equation of the circle is $(x+3)^2 + (y-2)^2 = 25$.
The centre of the circle is $(-3, 2)$.

2. Calculate the midpoint of $PQ$

Midpoint of $PQ = \left( \frac{-8 + 1}{2}, \frac{2 + (-1)}{2} \right)$
$= (-3.5, 0.5)$

3. Compare the midpoint with the centre to conclude

Since the midpoint of $PQ$ $(-3.5, 0.5)$ is not equal to the centre of the circle $(-3, 2)$, the line segment $PQ$ does not pass through the centre.
Therefore, $PQ$ is not a diameter.

1. State the centre of the circle from part (i)

From part (i), the centre of the circle $C$ is $(-3, 2)$.

2. Calculate the gradient of the normal (radius) at $R$

Gradient of the radius connecting $(-3, 2)$ to $R(0, 6)$:
$m_{\text{normal}} = \frac{6 - 2}{0 - (-3)} = \frac{4}{3}$

3. Find the gradient of the tangent at $R$

Since the tangent is perpendicular to the radius at the point of contact:
$m_{\text{tangent}} = -\frac{1}{m_{\text{normal}}} = -\frac{3}{4}$

4. Find the equation of the tangent line

Using the point $R(0, 6)$ and $m = -\frac{3}{4}$:
$y - 6 = -\frac{3}{4}(x - 0)$
$y = -\frac{3}{4}x + 6$

1. Find the centre of the original circle and the point of tangency

From part (i), the centre of $C$ is $(-3, 2)$ and its radius is $5$.
The point of tangency is $R(0, 6)$.

2. Determine the centre of the reflected circle

Since the circle is reflected in its tangent at $R$, the new circle will also have radius $5$ and touch the tangent at $R$. The new centre $C'$ lies on the normal line (the line passing through the original centre and $R$) such that $R$ is the midpoint of the original centre and $C'$.
Let the original centre be $C_0(-3, 2)$.
Using the midpoint formula:
$\frac{x + (-3)}{2} = 0 \implies x = 3$
$\frac{y + 2}{2} = 6 \implies y = 10$
The new centre is $(3, 10)$.

3. Write down the equation of the new circle

The radius remains $5$.
The equation of the new circle is:
$(x - 3)^2 + (y - 10)^2 = 5^2$
$(x - 3)^2 + (y - 10)^2 = 25$