已知圆 $C_1$ 的方程为 $x^2 - 6x + y^2 + 10y = 66$。

1. Find the center and radius of the circle by completing the square.

$x^2 - 6x + y^2 + 10y = 66$
$(x - 3)^2 - 9 + (y + 5)^2 - 25 = 66$
$(x - 3)^2 + (y + 5)^2 = 100$

The center of $C_1$ is $(3, -5)$ and the radius is $\sqrt{100} = 10$.

2. Compare the distance from the center to the axes with the radius to determine if they are tangents.

The distance from the center $(3, -5)$ to the $x$-axis is $|-5| = 5$.
Since $5 \neq 10$ (the radius), the $x$-axis is not a tangent.

The distance from the center $(3, -5)$ to the $y$-axis is $|3| = 3$.
Since $3 \neq 10$ (the radius), the $y$-axis is not a tangent.

1. Find the center and radius of the circle $C_1$ by completing the square.

$x^2 - 6x + y^2 + 10y = 66$
$(x - 3)^2 - 9 + (y + 5)^2 - 25 = 66$
$(x - 3)^2 + (y + 5)^2 = 100$
The center of $C_1$ is $(3, -5)$ and the radius is $\sqrt{100} = 10$.

2. Calculate the distance from the center to the point $(2, -4)$ and compare it with the radius.

Distance from $(3, -5)$ to $(2, -4)$:
$d = \sqrt{(2 - 3)^2 + (-4 - (-5))^2}$
$d = \sqrt{(-1)^2 + (1)^2} = \sqrt{2}$
Since $\sqrt{2} < 10$, the distance from the center to the point is less than the radius.

3. State the final conclusion.

Therefore, the point $(2, -4)$ lies inside $C_1$.

1. Apply the reflection transformation to the equation of $C_1$

To reflect a curve across the $y$-axis, we replace $x$ with $-x$ in its equation.
Equation of $C_2$:
$(-x)^2 - 6(-x) + y^2 + 10y = 66$

2. Simplify the equation

$x^2 + 6x + y^2 + 10y = 66$

(Alternatively, using the centre and radius: the centre of $C_1$ is $(3, -5)$ and its radius is $10$. Reflecting the centre across the $y$-axis gives $(-3, -5)$. The equation of $C_2$ is $(x + 3)^2 + (y + 5)^2 = 100$, which expands to the same result.)