A circle, $C$, has equation $x^2 + y^2 - 6x + 4y - 12 = 0$.

1. Group the $x$ and $y$ terms and complete the square for both.

$x^2 - 6x + y^2 + 4y = 12$
$(x - 3)^2 - 9 + (y + 2)^2 - 4 = 12$

2. Rearrange into the standard circle equation form to find the centre and radius.

$(x - 3)^2 + (y + 2)^2 = 12 + 9 + 4$
$(x - 3)^2 + (y + 2)^2 = 25$
$(x - 3)^2 + (y + 2)^2 = 5^2$

3. State the coordinates of the centre and the radius.

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

1. State the property of the normal to a circle and use the centre from part (i)

The normal to a circle at any point on its circumference must pass through the centre of the circle.
From (i), the centre of the circle $C$ is $(3, -2)$.

2. Substitute the coordinates of the centre into the equation of the normal to find $m$

Substitute $x = 3$ and $y = -2$ into $3y = m - 4x$:
$3(-2) = m - 4(3)$
$-6 = m - 12$
$m = 6$

1. Find the value of $m$ from part (ii) by substituting the centre of the circle into the normal equation

From (i), the centre of the circle is $(3, -2)$.
The normal to the circle at any point passes through the centre.
Substitute $(3, -2)$ into $3y = m - 4x$:
$3(-2) = m - 4(3)$
$-6 = m - 12$
$m = 6$
The equation of the normal is $3y = 6 - 4x$, or $y = 2 - \frac{4}{3}x$.

2. Find the coordinates of the possible points for $A$ by solving the normal and circle equations simultaneously

Substitute $y = 2 - \frac{4}{3}x$ into the circle equation $x^2 + y^2 - 6x + 4y - 12 = 0$:
$x^2 + \left(2 - \frac{4}{3}x\right)^2 - 6x + 4\left(2 - \frac{4}{3}x\right) - 12 = 0$
$x^2 + 4 - \frac{16}{3}x + \frac{16}{9}x^2 - 6x + 8 - \frac{16}{3}x - 12 = 0$
$\frac{25}{9}x^2 - \frac{50}{3}x = 0$
Multiply by 9:
$25x^2 - 150x = 0$
$25x(x - 6) = 0$
$x = 0$ or $x = 6$
When $x = 0$, $y = 2 - 0 = 2$. Point is $(0, 2)$.
When $x = 6$, $y = 2 - \frac{4}{3}(6) = -6$. Point is $(6, -6)$.

3. Determine which point is $A$ by checking the $y$-intercept of the tangent

The gradient of the normal is $-\frac{4}{3}$, so the gradient of the tangent is $\frac{3}{4}$.
For the point $(0, 2)$, the tangent equation is $y - 2 = \frac{3}{4}(x - 0) \implies y = \frac{3}{4}x + 2$.
The $y$-intercept is $2$, which is positive.
For the point $(6, -6)$, the tangent equation is $y - (-6) = \frac{3}{4}(x - 6) \implies y = \frac{3}{4}x - 10.5$.
The $y$-intercept is $-10.5$, which is negative.
Since the tangent at $A$ cuts the positive $y$-axis, $A$ must be $(0, 2)$.

1. Find the coordinates of $B$ using the fact that $AB$ is a diameter.

Since the tangents at $A$ and $B$ are parallel, $AB$ must be a diameter of the circle. Thus, the centre of the circle $(3, -2)$ is the midpoint of $A(0, 2)$ and $B(x, y)$.

$\frac{0 + x}{2} = 3 \implies x = 6$
$\frac{2 + y}{2} = -2 \implies y = -6$

So, $B$ is $(6, -6)$.

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

The gradient of the normal $AC$ is $\frac{2 - (-2)}{0 - 3} = -\frac{4}{3}$.
The tangent is perpendicular to the normal, so the gradient of the tangent at $A$ is $\frac{3}{4}$.
Since the tangent at $B$ is parallel to the tangent at $A$, its gradient is also $\frac{3}{4}$.

3. Find the equation of the tangent at $B$.

Using the point $B(6, -6)$ and gradient $m = \frac{3}{4}$:
$y - (-6) = \frac{3}{4}(x - 6)$
$y + 6 = \frac{3}{4}x - \frac{9}{2}$
$y = \frac{3}{4}x - \frac{21}{2}$