One, (20 points) Given the quadratic function
$$
y=x^{2}+2 m x-n^{2} \text {. }
$$
(1) 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 $P$, find the minimum value of $P$;
(2) If $m, n$ vary, these functions represent different parabolas. If each parabola intersects the coordinate axes at three distinct points, prove that the circles passing through these two intersection points all pass through the same fixed point, and find the coordinates of this fixed point.
(1) From the quadratic function passing through the point $(1,1)$, we get $m=\frac{n^{2}}{2}$.
Notice that
$$
\begin{array}{l}
m-(n+4)=\frac{n^{2}}{2}-(n+4) \\
=\frac{1}{2}\left(n^{2}-2 n-8\right)=\frac{1}{2}(n-4)(n+2),
\end{array}
$$

Therefore, $P=\left\{\begin{array}{ll}\frac{n^{2}}{2}, & n \leqslant-2 \text { or } n \geqslant 4 ; \\ n+4, & -2<n<4 .\end{array}\right.$
Using the function graph, we can see that when $n=-2$, $P_{\min }=2$.
(2) 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, so $x_{1} x_{2}=-n^{2}<0$. Therefore, $x_{1} 、 x_{2}$ are on opposite sides of the origin.
Also, $\left|x_{1} x_{2}\right|=n^{2} \times 1$, so there exists a point $P_{0}(0,1)$ such that $|O A| \cdot|O B|=\left|O P_{0}\right| \cdot|O C|$.
Thus, points $A 、 B 、 C 、 P_{0}$ are concyclic, meaning these circles must pass through the fixed point $P_{0}(0,1)$.