Example 6 Given the sets
$$
\begin{array}{l}
M=\left\{(x, y) \left\lvert\, y \geqslant \frac{1}{4} x^{2}\right.\right\}, \\
N=\left\{(x, y) \left\lvert\, y \leqslant-\frac{1}{4} x^{2}+x+7\right.\right\}, \\
D_{r}\left(x_{0}, y_{0}\right)=\left\{(x, y)!\left(x-x_{0}\right)^{2}+\left(y-y_{0}\right)^{2} \leqslant r^{2}\right\} .
\end{array}
$$

Find the largest $r$ such that $D_{r}\left(x_{0}, y_{0}\right) \subset M \cap N$.
[4]
Solution: Clearly, $M \cap N \neq \varnothing$.
As shown in Figure 6, in the Cartesian coordinate system,
$$
y=\frac{1}{4} x^{2}, y=-\frac{1}{4} x^{2}+x+7
$$

have vertices at $O(0,0)$ and $A(2,8)$, respectively, and the two parabolas are symmetric about point $B(1,4)$.

Therefore, the figure $T$ corresponding to $M \cap N$ in the Cartesian coordinate system is symmetric about point $B$, and the largest radius circle contained in $T$ should be the incircle of $T$.
We now prove that the center of this circle is $B$.
Otherwise, let the center of the circle be $B_{1}$, with radius $r$.
Take the point $B_{1}$ symmetric to $B$ as $B_{2}$. Then the circle with $B_{2}$ as the center and $r$ as the radius is also an incircle of $T$.
Draw the external common tangents of the two circles.
Since $T$ is convex, the region enclosed by the above two external common tangents and the two circles is inside $T$.

Therefore, the circle with $B$ as the center and $r$ as the radius is inside $T$ (excluding the boundary). Thus, the radius of the incircle of $T$ with $B$ as the center is greater than $r$, which is a contradiction.

Since $T$ is symmetric about point $B$, the incircle of $T$, $\odot B$, is the circle tangent to $y=\frac{1}{4} x^{2}$, denoted as $\odot B$, with radius $r_{0}$.
Let $\odot B$ be tangent to $y=\frac{1}{4} x^{2}$ at point $C(a, b)$. Then
$$
b=\frac{1}{4} a^{2} \text {. }
$$

The equation of the tangent line $l$ to $y=\frac{1}{4} x^{2}$ at point $C$ is
$$
y=\frac{a}{2}(x-a)+b .
$$

Clearly, $B C \perp l$, i.e.,
$$
\frac{b-4}{a-1} \cdot \frac{a}{2}=-1 \text {. }
$$

Substituting equation (1) into equation (2) and simplifying, we get
$$
a^{3}-8 a-8=0 \text {. }
$$

Solving this, we get $a=-2$ or $1 \pm \sqrt{5}$ (negative roots are discarded).
Also, $r_{0}=|B C|=\sqrt{(1-a)^{2}+\left(4-\frac{a^{2}}{4}\right)^{2}}$
$$
=\sqrt{\frac{25-5 \sqrt{5}}{2}} \text {, }
$$

Thus, $r_{\max }=r_{0}=\sqrt{\frac{25-5 \sqrt{5}}{2}}$.