圆 $C_1$ 的圆心在直线 $2x + y = 0$ 上，且通过点 $P(3, -1)$ 和 $Q(-4, -2)$。

1. Find the coordinates of the midpoint of $PQ$

Midpoint of $PQ = \left( \frac{3 + (-4)}{2}, \frac{-1 + (-2)}{2} \right) = \left( -\frac{1}{2}, -\frac{3}{2} \right)$

2. Find the gradient of $PQ$ and the gradient of its perpendicular bisector

Gradient of $PQ = \frac{-2 - (-1)}{-4 - 3} = \frac{-1}{-7} = \frac{1}{7}$

Gradient of perpendicular bisector = $-\frac{1}{\frac{1}{7}} = -7$

3. Find the equation of the perpendicular bisector

Equation of perpendicular bisector:
$y - \left(-\frac{3}{2}\right) = -7\left(x - \left(-\frac{1}{2}\right)\right)$
$y + \frac{3}{2} = -7x - \frac{7}{2}$
$y = -7x - 5$

1. Find the equation of the perpendicular bisector of $PQ$ (from part (i)).

Midpoint of $PQ = \left(\frac{3 + (-4)}{2}, \frac{-1 + (-2)}{2}\right) = (-0.5, -1.5)$
Gradient of $PQ = \frac{-2 - (-1)}{-4 - 3} = \frac{1}{7}$
Gradient of perpendicular bisector = $-7$
Equation of perpendicular bisector: $y - (-1.5) = -7(x - (-0.5)) \implies y = -7x - 5$

2. Find the coordinates of the centre of the circle.

The centre lies on the perpendicular bisector of $PQ$ and the line $2x + y = 0$.
Substitute $y = -2x$ into $y = -7x - 5$:
$-2x = -7x - 5$
$5x = -5 \implies x = -1$
$y = -2(-1) = 2$
The centre of $C_1$ is $(-1, 2)$.

3. Calculate the square of the radius.

The radius $r$ is the distance from the centre $(-1, 2)$ to $P(3, -1)$.
$r^2 = (3 - (-1))^2 + (-1 - 2)^2$
$r^2 = 4^2 + (-3)^2 = 16 + 9 = 25$

4. State the equation of the circle.

Equation of $C_1$: $(x - (-1))^2 + (y - 2)^2 = 25$
$(x + 1)^2 + (y - 2)^2 = 25$

1. State the centre and radius of $C_1$ from part (ii)

From part (ii), the centre of $C_1$ is $C(-1, 2)$ and its radius is $r = 5$.

2. Calculate the distance from the centre to the point $R(2, 5)$

Distance $CR = \sqrt{(2 - (-1))^2 + (5 - 2)^2}$
$CR = \sqrt{3^2 + 3^2} = \sqrt{18}$

3. Compare the distance to the radius to conclude

Since $\sqrt{18} < \sqrt{25}$, the distance $CR < r$.
Therefore, the point $R(2, 5)$ lies inside $C_1$.

1. Identify the centre and radius of $C_1$ from earlier parts

From part (ii), the centre of $C_1$ is $(-1, 2)$ and its radius squared is $r^2 = 25$.

2. Find the centre of $C_2$ by reflecting the centre of $C_1$ in the $y$-axis

When reflected in the $y$-axis, the $x$-coordinate of the centre changes sign while the $y$-coordinate remains the same.
Centre of $C_2 = (1, 2)$

3. Write the equation of $C_2$

The radius of $C_2$ is the same as $C_1$.
Equation of $C_2$: $(x - 1)^2 + (y - 2)^2 = 25$