Two points $A(-3, -6)$ and $B(7, -6)$ lie on a circle with centre $G$. The equation of the tangent to the circle at the point $Q(1, -12)$ is $x + 5y + 59 = 0$.

1. Find the equation of the perpendicular bisector of chord $AB$.

The points $A(-3, -6)$ and $B(7, -6)$ have the same $y$-coordinate, so the chord $AB$ is horizontal.
The midpoint of $AB$ is $\left(\frac{-3+7}{2}, -6\right) = (2, -6)$.
The perpendicular bisector of $AB$ is a vertical line passing through the midpoint, so its equation is $x = 2$.

2. Find the gradient of the normal to the circle at $Q$.

The tangent at $Q(1, -12)$ is $x + 5y + 59 = 0$, which can be written as $5y = -x - 59 \implies y = -\frac{1}{5}x - \frac{59}{5}$.
The gradient of the tangent is $m_t = -\frac{1}{5}$.
The normal to the circle at $Q$ passes through the centre $G$. Its gradient is $m_n = -\frac{1}{m_t} = 5$.

3. Find the equation of the normal at $Q$.

Using the point $Q(1, -12)$ and gradient $m_n = 5$:
$y - (-12) = 5(x - 1)$
$y + 12 = 5x - 5$
$y = 5x - 17$

4. Find the coordinates of the centre $G$ by intersecting the perpendicular bisector and the normal.

The centre $G$ lies on both the perpendicular bisector of $AB$ ($x = 2$) and the normal at $Q$ ($y = 5x - 17$).
Substitute $x = 2$ into the normal equation:
$y = 5(2) - 17 = 10 - 17 = -7$
Thus, the coordinates of $G$ are $(2, -7)$. (Shown)

1. Calculate the square of the radius using the centre $G$ and a point on the circle

Using centre $G(2, -7)$ and point $A(-3, -6)$:
$r^2 = (-3 - 2)^2 + (-6 - (-7))^2$
$r^2 = (-5)^2 + (1)^2 = 25 + 1 = 26$

2. Write the equation of the circle in standard form and expand it

Equation of the circle:
$(x - 2)^2 + (y + 7)^2 = 26$
$x^2 - 4x + 4 + y^2 + 14y + 49 = 26$
$x^2 + y^2 - 4x + 14y + 53 - 26 = 0$
$x^2 + y^2 - 4x + 14y + 27 = 0$

3. Compare with the general form to state $f$, $g$, and $c$

Comparing $x^2 + y^2 - 4x + 14y + 27 = 0$ with $x^2 + y^2 + 2gx + 2fy + c = 0$:
$2g = -4 \implies g = -2$
$2f = 14 \implies f = 7$
$c = 27$