圆的方程为 $x^2 + y^2 - 8x + 4y - 16 = 0$。

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

$x^2 - 8x + y^2 + 4y = 16$

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

$(x - 4)^2 - 16 + (y + 2)^2 - 4 = 16$
$(x - 4)^2 + (y + 2)^2 = 16 + 16 + 4$
$(x - 4)^2 + (y + 2)^2 = 36$

3. Identify the centre and radius from the standard form

Centre $C = (4, -2)$
Radius $r = \sqrt{36} = 6$

1. Substitute the linear equation into the circle equation

Substitute $y = 2x + 2$ into $x^2 + y^2 - 8x + 4y - 16 = 0$:
$x^2 + (2x + 2)^2 - 8x + 4(2x + 2) - 16 = 0$

2. Expand and simplify to form a quadratic equation in $x$

$x^2 + 4x^2 + 8x + 4 - 8x + 8x + 8 - 16 = 0$
$5x^2 + 8x - 4 = 0$

3. Solve the quadratic equation for $x$

$(5x - 2)(x + 2) = 0$
$x = \frac{2}{5} \text{ or } x = -2$

4. Find the corresponding $y$-coordinates and state the points

When $x = \frac{2}{5} = 0.4$:
$y = 2(0.4) + 2 = 2.8$
When $x = -2$:
$y = 2(-2) + 2 = -2$
The coordinates are $(-2, -2)$ and $(0.4, 2.8)$.

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

The original circle is $x^2 + y^2 - 8x + 4y - 16 = 0$.
Completing the square:
$(x - 4)^2 - 16 + (y + 2)^2 - 4 - 16 = 0$
$(x - 4)^2 + (y + 2)^2 = 36$
The centre $C$ is $(4, -2)$ and the radius $r = \sqrt{36} = 6$.

2. Determine the radius of the new circle

Since the area of the new circle is 4 times the area of the original circle, the ratio of their areas is $4$.
The ratio of their radii is $\sqrt{4} = 2$.
New radius $R = 2 \times 6 = 12$.

3. Write the equation of the new circle

Using the centre $C(4, -2)$ and new radius $R = 12$:
$(x - 4)^2 + (y + 2)^2 = 12^2$
$(x - 4)^2 + (y + 2)^2 = 144$