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

Find the largest $r$ such that $D_{r}\left(x_{0}, y_{0}\right) \subset M \cap N$.
2. Let $f(x)=\frac{1}{4} x^{2}$,
$$
g(x)=-\frac{1}{4} x^{2}+x+7=-\frac{1}{4}(x-2)^{2}+8 \text {. }
$$

Then $M \cap N \neq \varnothing \Leftrightarrow f(x) \leqslant g(x)$ has real solutions $\Leftrightarrow x^{2}-2 x-14 \leqslant 0$ has real solutions.
Thus, $M \cap N \neq \varnothing$.
In the Cartesian coordinate system, the vertices of $y=f(x)$ and $y=g(x)$ are $O(0,0)$ and $A(2,8)$, respectively, and the two parabolas are symmetric about the point $B(1,4)$.

Therefore, the figure $T$ corresponding to $M \cap N$ in the coordinate system is symmetric about the point $B$, and the largest radius circle contained in $T$ should be the incircle of $T$.
We now prove: the center of this circle is $B$.
Otherwise, let the center of this circle be $B_{1}$ (not coinciding with point $B$), with radius $r$. Take the point $B_{2}$, which is the symmetric point of $B_{1}$ about $B$. Then the circle with center $B_{2}$ and radius $r$ is also an incircle of the figure $T$.
Draw two external common tangents to the two circles.
Since the figure $T$ is convex, the region enclosed by the above two external common tangents and the two circles is inside the figure $T$.

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

Since the figure $T$ is symmetric about the point $B$, the incircle $\odot B$ is the circle with center $B$ that is internally tangent to $y=\frac{1}{4} x^{2}$. Let its radius be $r_{0}$.
Suppose $\odot B$ is tangent to $y=\frac{1}{4} x^{2}$ at point $C(a, b)$. Then $b=\frac{a^{2}}{4}$.
The tangent line $l$ to $y=\frac{1}{4} x^{2}$ at point $C$ is given by
$$
y=\frac{a}{2}(x-a)+b .
$$

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

Substituting (1) into (2) and simplifying, we get $a^{3}-8 a-8=0$.
Solving, we get $a=-2$ or $1 \pm \sqrt{5}$.
$$
\text { Also, } r_{0}=|B C|=\sqrt{\left(\frac{a}{2}\right)^{2}+1} \cdot|1-a| \text {, substituting the value of } a
$$

we find that the minimum value of $r_{0}$ is the radius of $\odot B$, i.e., when $a=1+\sqrt{5}$, $r_{0}=\sqrt{\frac{25-5 \sqrt{5}}{2}}$.
Therefore, $r_{\max }=r_{0}=\sqrt{\frac{25-5 \sqrt{5}}{2}}$.