Four. (18 points) Given the quadratic function
$$
y=x^{2}-2 m x-2 m^{2}(m \neq 0)
$$

whose graph intersects the $x$-axis at points $A$ and $B$, and its vertex lies on the circle with $A B$ as its diameter.
(1) Prove: $A$ and $B$ are two distinct intersection points on the $x$-axis;
(2) Find the analytical expression of the quadratic function;
(3) Suppose the circle with $A B$ as its diameter intersects the $y$-axis at points $C$ and $D$, find the length of the chord $C D$.
(1) Since $\Delta=4 m^{2}+8 m^{2}=12 m^{2}>0$, therefore, $A$ and $B$ are two distinct intersection points on the $x$-axis.
(2) Construct the parabola
$$
y=x^{2}-2 m x-2 m^{2}
$$
(see Figure 11). Let the intersection points with the $x$-axis be $A\left(x_{1}, 0\right)$ and
$$
B\left(x_{2}, 0\right)\left(x_{1}<x_{2}\right) \text {. }
$$

Draw a circle with $AB$ as its diameter, with the center at $P$,
$\odot P$ is tangent to the parabola at the vertex $Q$. Then
$$
\begin{array}{l}
|A B|=\left|x_{1}-x_{2}\right|=\sqrt{\left(x_{1}-x_{2}\right)^{2}} \\
\Rightarrow|A B|=\sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}} .
\end{array}
$$

Since $x_{1}$ and $x_{2}$ are the roots of the equation $x^{2}-2 m x-2 m^{2}=0$, we have
$$
x_{1}+x_{2}=2 m, x_{1} x_{2}=-2 m^{2} .
$$

Thus, $|A B|=\sqrt{4 m^{2}+8 m^{2}}=2 \sqrt{3}|m|$.
Therefore, the radius of $\odot P$ is $P Q=|\sqrt{3} m|$.
The vertex of the parabola $y=x^{2}-2 m x-2 m^{2}$ is $Q\left(m,-3 m^{2}\right)$.
By the given conditions, $|\sqrt{3} m|=\left|-3 m^{2}\right|$.
Solving, we get $m=0$ (discard), $|m|=\frac{\sqrt{3}}{3} \Rightarrow m= \pm \frac{\sqrt{3}}{3}$.
Thus, the quadratic function is
$y=x^{2}-\frac{2 \sqrt{3}}{3} x-\frac{2}{3}$ or $y=x^{2}+\frac{2 \sqrt{3}}{3} x-\frac{2}{3}$.
(3) From $y=x^{2}-\frac{2 \sqrt{3}}{3} x-\frac{2}{3}$, setting $y=0$, we get
$$
x_{1}=\frac{\sqrt{3}-3}{3}, x_{2}=\frac{\sqrt{3}+3}{3} \text {. }
$$

Since $\triangle A B C$ is a right triangle, we have
$$
|O C|^{2}=|O A| \cdot|O B| \text {. }
$$

Thus, $|O C|^{2}=\left|x_{1}\right| \cdot\left|x_{2}\right|$
$$
=\left|\frac{\sqrt{3}-3}{3}\right| \cdot\left|\frac{\sqrt{3}+3}{3}\right|=\frac{2}{3} \text {. }
$$

Therefore, $|O C|=\sqrt{\frac{2}{3}}=\frac{\sqrt{6}}{3}$.
Hence, the chord $|C D|=\frac{2 \sqrt{6}}{3}$.