11. Given $a>0$, the curve $C:|y|=1-a x^{2}\left(|x| \leqslant \frac{1}{\sqrt{a}}\right)$ can contain a circle of radius $r$, and when the center of the circle is $(0,0)$, the maximum value of $r$ is $f(a)$.
(1) Find the expression for $f(a)$;
(2) Prove that when the center of the circle is not at $(0,0)$, $r \leqslant f(a)$.
11. Solution: (1) From $|x| \leqslant \frac{1}{\sqrt{a}}, 1-a x^{2} \geqslant 0$, for any point $\left(x_{0}, y_{0}\right)$ on the curve $C$, we know: $r^{2} \leqslant x_{0}^{2}+y_{0}^{2}=x_{0}^{2}+\left(1-a x_{0}^{2}\right)^{2}=a^{2} x_{0}^{4}-(2 a-1) x_{0}^{2}+1$. (※)
(i) When $0\frac{1}{2}$, $a^{2} x_{0}^{4}-(2 a-1) x_{0}^{2}+1=\left(a x_{0}^{2}-\frac{2 a-1}{2 a}\right)^{2}+\frac{4 a-1}{4 a^{2}}$ $\geqslant \frac{4 a-1}{4 a^{2}}$ (equality holds when $x_{0}= \pm \frac{\sqrt{4 a-2}}{2 a}$),
Also, when $x_{0}^{2}=\frac{2 a-1}{2 a^{2}}$, we get $r^{2} \leqslant \frac{4 a-1}{4 a^{2}}$, so $f(a)=\frac{\sqrt{4 a-1}}{2 a}$. Therefore, $f(a)=\left\{\begin{array}{c}1\left(0\frac{1}{2}\right)\end{array}\right.$.
(2) Let the center of the circle be $P(p, q)$. Since the curve $C$ is symmetric about the coordinate axes, we can assume $p, q \geqslant 0$, and denote the part of the curve $C$ where $x, y \geqslant 0$ as $C^{*}$.
(i) When $0\frac{1}{2}$, as shown in Figure (2), take a point $T\left(\frac{\sqrt{4 a-2}}{2 a}, \frac{1}{2 a}\right)$ on $C^{*}$, connect $O T$, and draw lines $l_{1}, l_{2}$ through $T$ parallel to the $x$-axis and $y$-axis, respectively. Since $C^{*}$ is a segment of the quadratic function $y=-a x^{2}+1$ to the right of the axis of symmetry, by the monotonicity of the function, we know: any point on $C^{*}$ to the right of $T$ is below $T$, and any point on $C^{*}$ to the left of $T$ is above $T$. Also, from the solution in (1), we know $|O T|=f(a)$.
(1) When point $P$ is on $O T$, $r \leqslant|P T| \leqslant|O T|=f(a)$;
(2) When point $P$ is in $\angle T O x$ (including the $x$-axis), draw $P M / / O T$ intersecting $C^{*}$, the $x$-axis, and the $y$-axis at $M$, $E$, and $F$, respectively, then $r \leqslant|P M|<|E F|=|O T|=f(a)$.
(3) When point $P$ is in $\angle T O y$ (including the $y$-axis), draw $P M / / O T$ intersecting $C^{*}$, the $y$-axis, and the $x$-axis at $M$, $G$, and $H$, respectively, then $r \leqslant|P M|<|G H|=|O T|=f(a)$. In summary, when the center is $(p, q)$, $r \leqslant f(a)$.