9. Given that $\odot O^{\prime}$ passes through the fixed point $A(0, p) (p>0)$, the center $O^{\prime}$ moves on the parabola $x^{2}=2 p y$, and $M N$ is the chord of the circle $O^{\prime}$ intercepted by the $x$-axis.
(1) When point $O^{\prime}$ moves, does $|M N|$ change? Prove your conclusion;
(2) When $|O A|$ is the arithmetic mean of $|O M|$ and $|O N|$, determine the position relationship between the directrix of the parabola $C$ and the circle $O^{\prime}$, and explain your reasoning.
9. Solution: (1) Let $O^{\prime}\left(x_{0}, y_{0}\right)$, then $x_{0}^{2}=2 p y_{0}\left(y_{0} \geqslant 0\right)$, so the radius of circle $O^{\prime}$, $\left|O^{\prime} A\right|=\sqrt{x_{0}^{2}+\left(y_{0}-p\right)^{2}}$. The equation of $\odot O^{\prime}$ is $\left(x-x_{0}\right)^{2}+\left(y-y_{0}\right)^{2}=x_{0}^{2}+\left(y_{0}-p\right)^{2}$. Setting $y=0$ and substituting $x_{0}^{2}=2 p y_{0}$, we get $x^{2}-2 x_{0} x+x_{0}^{2}-p^{2}=0$,
solving for $x_{1}=x_{0}-p, x_{2}=x_{0}+p, \therefore|M N|=\left|x_{1}-x_{2}\right|=2 p, \therefore|M N|$ is constant, with a value of $2 p$.
(2) Assume $M\left(x_{0}-p, 0\right), N\left(x_{0}+p, 0\right)$,

From the problem, $2|O A|=|O M|+|O N|$ gives $2 p=\left|x_{0}-p\right|+\left|x_{0}+p\right|$.
$\therefore-p \leqslant x_{0} \leqslant p, \therefore O^{\prime}$ to the directrix of the parabola $y=-\frac{p}{2}$, the distance $\quad d=y_{0}+\frac{p}{2}=\frac{x_{0}^{2}+p^{2}}{2 p}$.
The radius of $\odot O^{\prime}$, $\left|O^{\prime} A\right|=\sqrt{x_{0}^{2}+\left(y_{0}-p\right)^{2}}=\sqrt{x_{0}^{2}+\left(\frac{x_{0}^{2}}{2 p}-p\right)^{2}}=\frac{1}{2 p} \sqrt{x_{0}^{4}+4 p^{4}}$.
$$
\therefore r>d \Leftrightarrow x_{0}^{4}+4 p^{4}>\left(x_{0}^{2}+p^{2}\right)^{2} \Leftrightarrow x_{0}^{2}0) \text {. }
$$

Thus, $r>d$, meaning $\odot O^{\prime}$ always intersects the directrix of the parabola.