9. Given the quadratic function $y=x^{2}+2 m x-n^{2}$.
(1) If $m, n$ vary, their graphs are different parabolas. If each parabola intersects the coordinate axes at three different points, and a circle is drawn through these three points, prove that all these circles pass through the same fixed point, and find the coordinates of this fixed point.
(2) If the graph of this quadratic function passes through the point $(1,1)$, and let the larger of the two numbers $m, n+4$ be $P$, try to find the minimum value of $P$.
9. Solution: (1) The graph intersects the coordinate axes at three different points, which can be set as $A\left(x_{1}, 0\right), B\left(x_{2}, 0\right), C\left(0,-n^{2}\right)$.

Also, $x_{1} x_{2}=-n^{2}$. If $n=0$, it would contradict the condition of three intersection points, $\therefore x_{1} x_{2}=-n^{2}<0, \therefore x_{1}, x_{2}$ are on opposite sides of the origin. Since $\left|x_{1} x_{2}\right|=n^{2} \cdot 1, \therefore$ there exists a point $P_{0}(0,1)$ such that $|O A| \cdot|O B|=\left|O P_{0}\right| \cdot|O C|$, so $A, B, C, P_{0}$ are concyclic, $\therefore$ these parabolas must pass through the fixed point $P_{0}(0,1)$.
(2) From passing through the point (1,1), we get: $m=\frac{n^{2}}{2}$.
To compare the sizes of $m$ and $n+4$, i.e., $m-(n+4)=\frac{n^{2}}{2}-(n+4)=\frac{1}{2}\left(n^{2}-2 n-8\right)=\frac{1}{2}(n-4)(n+2)$.
$$
\therefore P=\left\{\begin{array}{ll}
\frac{n^{2}}{2} & (n \leqslant-2 \text { or } n \geqslant 4) \\
n+4 & (-2<n<4)
\end{array}\right.
$$

As shown in the figure, when $n=-2$, $P_{\text {min }}=2$.