点 $A(5, 7)$ 和 $B(-11, 15)$ 在圆 $C$ 上。圆心 $O$ 在直线 $y = -2x - 3$ 上。

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

Midpoint of $AB = \left( \frac{5 + (-11)}{2}, \frac{7 + 15}{2} \right) = \left( \frac{-6}{2}, \frac{22}{2} \right) = (-3, 11)$

2. Find the gradient of $AB$ and the gradient of the perpendicular bisector.

Gradient of $AB = \frac{15 - 7}{-11 - 5} = \frac{8}{-16} = -\frac{1}{2}$

Gradient of perpendicular bisector $= -\frac{1}{-\frac{1}{2}} = 2$

3. Find the equation of the perpendicular bisector.

Using $y - y_1 = m(x - x_1)$ with the midpoint $(-3, 11)$ and gradient $2$:
$y - 11 = 2(x - (-3))$
$y - 11 = 2(x + 3)$
$y = 2x + 6 + 11$
$y = 2x + 17$

1. Find the equation of the perpendicular bisector of $AB$ (result from part a)

Midpoint of $AB = \left(\frac{5-11}{2}, \frac{7+15}{2}\right) = (-3, 11)$
Gradient of $AB = \frac{15-7}{-11-5} = -\frac{1}{2}$
Gradient of perpendicular bisector = $2$
Equation: $y - 11 = 2(x + 3) \implies y = 2x + 17$

2. Find the coordinates of the centre of the circle

The centre lies on the perpendicular bisector of any chord, so it lies on $y = 2x + 17$. It is also given that the centre lies on $y = -2x - 3$.
Solving simultaneously:
$2x + 17 = -2x - 3$
$4x = -20 \implies x = -5$
$y = 2(-5) + 17 = 7$
Centre of circle is $(-5, 7)$.

3. Calculate the radius of the circle

Radius $r$ is the distance from the centre $(-5, 7)$ to point $A(5, 7)$.
$r^2 = (5 - (-5))^2 + (7 - 7)^2$
$r^2 = 10^2 + 0 = 100$

4. Form the equation of the circle

Using the standard form $(x - a)^2 + (y - b)^2 = r^2$:
$(x - (-5))^2 + (y - 7)^2 = 100$
$(x + 5)^2 + (y - 7)^2 = 100$

1. Identify that $AD$ is a diameter of the circle

Since angle $ABD = 90^\circ$ and points $A$, $B$, and $D$ lie on the circle, by the converse of the angle in a semicircle theorem, $AD$ must be a diameter of the circle. Thus, the centre of the circle $O(-5, 7)$ is the midpoint of $AD$.

2. Use the midpoint formula to find the coordinates of $D$

Let the coordinates of $D$ be $(x, y)$.
Using the midpoint formula with $A(5, 7)$ and centre $O(-5, 7)$:
$\frac{5 + x}{2} = -5 \implies 5 + x = -10 \implies x = -15$
$\frac{7 + y}{2} = 7 \implies 7 + y = 14 \implies y = 7$

The coordinates of $D$ are $(-15, 7)$.

1. Set up the inequality for the intersection of the line and the circle

From part (b), the equation of the circle is $(x + 5)^2 + (y - 7)^2 = 100$.
Substitute $x = k$ into the equation:
$(k + 5)^2 + (y - 7)^2 = 100$
$(y - 7)^2 = 100 - (k + 5)^2$
For the line to intersect the circle at two points, there must be two distinct real values for $y$, so:
$100 - (k + 5)^2 > 0$

2. Solve the inequality to find the range of $k$

$(k + 5)^2 < 100$
$-10 < k + 5 < 10$
$-15 < k < 5$

1. Identify the centre and radius of the original circle $C$ from earlier parts

From part (b), the centre of $C$ is $(-5, 7)$ and its radius squared is $100$.

2. Find the centre of the reflected circle $C_2$

When reflected in the $y$-axis, the $x$-coordinate of the centre changes sign while the $y$-coordinate remains the same.
New centre $= (5, 7)$

3. Write the equation of $C_2$

The radius remains the same ($r^2 = 100$).
Equation of $C_2$: $(x - 5)^2 + (y - 7)^2 = 100$