Four. (16 points) Given the quadratic function
$$
y=a x^{2}+4 a x+4 a-1
$$

the graph of which is $C_{1}$.
(1) Find the function expression of the image $C_{2}$ that is centrally symmetric to $C_{1}$ about the point $R(1,0)$;
(2) Let the intersection points of curves $C_{1}$ and $C_{2}$ with the $y$-axis be $A$ and $B$, respectively. When $|A B|=18$, find the value of $a$.
(1) From $y=$ $a(x+2)^{2}-1$, we know that the vertex of parabola $C_{1}$ is $M(-2,-1)$.

From Figure 6, we know that the point $M(-2,-1)$ is centrally symmetric to point $N(4,1)$ with respect to point $R(1,0)$. With $N(4,1)$ as the vertex, the image $C_{2}$, which is centrally symmetric to parabola $C_{1}$ with respect to point $R(1,0)$, is also a parabola, and the opening directions of $C_{1}$ and $C_{2}$ are opposite. Therefore, the function expression of parabola $C_{2}$ is $y=-a(x-4)^{2}+1$, which is
$$
y=-a x^{2}+8 a x-16 a+1 \text {. }
$$
(2) Let $x=0$, we get the y-coordinates of the intersection points $A$ and $B$ of parabolas $C_{1}$ and $C_{2}$ with the y-axis are $4 a-1$ and $-16 a+1$, respectively, so
$$
|A B|=|(4 a-1)-(-16 a+1)|=|20 a-2| \text {. }
$$

Noting that $|20 a-2|=18$.
When $a \geqslant \frac{1}{10}$, we have $20 a-2=18$, which gives $a=1$;
When $a<\frac{1}{10}$, we have $2-20 a=18$, which gives $a=-\frac{4}{5}$.