The diagram below is not drawn to scale. In the diagram, D, E and F are points on the circle C1.

1. Calculate the gradient of the line segment FD.

Gradient of $FD = \frac{4 - 2}{7 - (-1)} = \frac{2}{8} = \frac{1}{4}$

2. Calculate the gradient of the line segment FE.

Gradient of $FE = \frac{-6 - 2}{1 - (-1)} = \frac{-8}{2} = -4$

3. Show that FD and FE are perpendicular.

Gradient of $FD \times$ Gradient of $FE = \frac{1}{4} \times (-4) = -1$
Since the product of their gradients is $-1$, $FD \perp FE$, which means $\angle DFE = 90^\circ$.

4. Conclude that DE is the diameter using circle properties.

Since $\angle DFE = 90^\circ$ and $D, E, F$ lie on the circle $C_1$, by the converse of the angle in a semicircle theorem, $DE$ must be the diameter of $C_1$.

5. Find the centre of the circle by calculating the midpoint of the diameter DE.

Centre of $C_1$ = Midpoint of $DE$
$= \left( \frac{7 + 1}{2}, \frac{4 + (-6)}{2} \right)$
$= \left( \frac{8}{2}, \frac{-2}{2} \right)$
$= (4, -1)$

1. Find the square of the radius of the circle using the centre from part (a) and point D

From part (a), the centre of $C_1$ is $(4, -1)$.
Using point $D(7, 4)$, the square of the radius $r^2$ is:
$r^2 = (7 - 4)^2 + (4 - (-1))^2$
$r^2 = 3^2 + 5^2 = 9 + 25 = 34$

2. Write the equation of the circle in standard form

The equation of the circle is:
$(x - 4)^2 + (y + 1)^2 = 34$

3. Expand the equation to the required general form

$x^2 - 8x + 16 + y^2 + 2y + 1 = 34$
$x^2 + y^2 - 8x + 2y + 17 - 34 = 0$
$x^2 + y^2 - 8x + 2y - 17 = 0$

1. Find the centre of circle $C_2$ by reflecting the centre of $C_1$ in the line $x = -2$.

From part (a), the centre of $C_1$ is $(4, -1)$.
The line of reflection is $x = -2$.
The distance from the $x$-coordinate of the centre to the line of reflection is $4 - (-2) = 6$.
The $x$-coordinate of the reflected centre is $-2 - 6 = -8$.
The $y$-coordinate remains unchanged.
Centre of $C_2$ is $(-8, -1)$.

2. Write the equation of $C_2$ using the new centre and the same radius.

From part (a), the radius squared of $C_1$ is $r^2 = (7-4)^2 + (4 - (-1))^2 = 34$.
Since reflection preserves the size of the circle, the radius of $C_2$ is also $\sqrt{34}$.
The equation of $C_2$ is:
$(x - (-8))^2 + (y - (-1))^2 = 34$
$(x + 8)^2 + (y + 1)^2 = 34$
Expanding this gives:
$x^2 + 16x + 64 + y^2 + 2y + 1 = 34$
$x^2 + y^2 + 16x + 2y + 31 = 0$

1. Recall or calculate the centre and radius of $C_1$ and $C_2$ from earlier parts

From (a), centre of $C_1$ is $M(4, -1)$ and radius $r = \sqrt{(7-4)^2 + (4-(-1))^2} = \sqrt{9+25} = \sqrt{34}$.
From (c), $C_2$ is the reflection of $C_1$ in $x = -2$. The centre of $C_2$ is $M'(-8, -1)$ and its radius is also $\sqrt{34}$.

2. Calculate the distance from the point (3, 4) to the centre of $C_1$

Let $P$ be $(3, 4)$.
Distance $PM = \sqrt{(3 - 4)^2 + (4 - (-1))^2} = \sqrt{(-1)^2 + 5^2} = \sqrt{26}$.
Since $\sqrt{26} < \sqrt{34}$, $P$ lies inside $C_1$.

3. Calculate the distance from the point (3, 4) to the centre of $C_2$

Distance $PM' = \sqrt{(3 - (-8))^2 + (4 - (-1))^2} = \sqrt{11^2 + 5^2} = \sqrt{121 + 25} = \sqrt{146}$.
Since $\sqrt{146} > \sqrt{34}$, $P$ lies outside $C_2$.