圆的方程为 $x^2 + y^2 + 2x - 8y - 8 = 0$。

1. Rearrange the equation by completing the square for $x$ and $y$.

$x^2 + 2x + y^2 - 8y = 8$
$(x + 1)^2 - 1^2 + (y - 4)^2 - (-4)^2 = 8$
$(x + 1)^2 - 1 + (y - 4)^2 - 16 = 8$
$(x + 1)^2 + (y - 4)^2 = 25$

2. Identify the coordinates of the centre.

Comparing with $(x - a)^2 + (y - b)^2 = r^2$,
Centre $C = (-1, 4)$

3. Identify the radius of the circle.

$r^2 = 25$
$r = 5$

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

From part (a), the original circle is $x^2 + y^2 + 2x - 8y - 8 = 0$.
Completing the square:
$(x + 1)^2 - 1 + (y - 4)^2 - 16 - 8 = 0$
$(x + 1)^2 + (y - 4)^2 = 25$
Centre $C = (-1, 4)$, Radius $r = 5$.

2. Find the coordinates of the reflected centre

The line of reflection is $y = 1$.
The $x$-coordinate of the centre remains unchanged: $x' = -1$.
The $y$-coordinate of the centre is reflected across $y = 1$. The distance from $y = 4$ to $y = 1$ is $3$ units.
The reflected $y$-coordinate is $1 - 3 = -2$.
New centre $C' = (-1, -2)$.

3. Write the equation of the reflected circle

The radius remains the same, $r = 5$.
The equation of the reflected circle is:
$(x - (-1))^2 + (y - (-2))^2 = 5^2$
$(x + 1)^2 + (y + 2)^2 = 25$
Expanding this gives:
$x^2 + 2x + 1 + y^2 + 4y + 4 = 25$
$x^2 + y^2 + 2x + 4y - 20 = 0$

1. Determine the diameter of the circle from its radius

From part (a), the radius of the circle is $r = \sqrt{(-1)^2 + 4^2 - (-8)} = 5$.
The diameter of the circle is $d = 2 \times 5 = 10$.

2. Calculate the area of the inscribed square

The biggest square that can be drawn inside the circle is inscribed in it, meaning its diagonal is equal to the diameter of the circle.
Let the side length of the square be $s$.
By Pythagoras' theorem, $s^2 + s^2 = d^2$
$2s^2 = 10^2 = 100$
$s^2 = 50$
The area of the square is $s^2 = 50$.