10. (20 points) Given the parabola $C: y^{2}=4 x$ with focus $F$, points $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$ lie on the parabola $C$, and $x_{1}+x_{2}=4$. Two circles are drawn with diameters $A F$ and $B F$ respectively. The common chord of these two circles lies on the line $l$, which intersects $C$ at points $M$ and $N$. Find the range of $|M N|$.
10. From the problem, we know the focus $F(1,0)$. Let the midpoints of $A F$ and $B F$ be $O_{1}$ and $O_{2}$, respectively.
(1) When points $A$, $B$, and $F$ are not collinear, the common chord is perpendicular to the line connecting the midpoints $O_{1} O_{2}$, and $O_{1} O_{2}$ is the median of $\triangle A B F$. Thus, $O_{1} O_{2} \parallel A B$, $M N \perp A B$, and $M N$ passes through point $F$.
Let $A\left(\frac{y_{1}^{2}}{4}, y_{1}\right), B\left(\frac{y_{2}^{2}}{4}, y_{2}\right)$.
If $y_{1}=-y_{2}$, in this case, the line $M N$ is the axis, intersecting the parabola at only one point $O$, which is discarded.
Therefore, $y_{1} \neq -y_{2}$, and the slope of $l_{A B}$ must exist, denoted as
$$
k=\frac{y_{1}-y_{2}}{\frac{y_{1}^{2}}{4}-\frac{y_{2}^{2}}{4}}=\frac{4}{y_{1}+y_{2}}.
$$

Since $\frac{y_{1}^{2}+y_{2}^{2}}{4}=4$, we have $32 \geqslant (y_{1}+y_{2})^{2}$.
Also, $y_{1} \neq y_{2}$, so $\left|y_{1}+y_{2}\right| \in (0, 4 \sqrt{2})$.
Thus, $k \in \left(-\infty, -\frac{\sqrt{2}}{2}\right) \cup \left(\frac{\sqrt{2}}{2}, +\infty\right)$.
Let $l_{M N}: y=-\frac{1}{k}(x-1)$, substituting into the parabola equation gives $y^{2}+4 k y-4=0$.
Thus, let $M\left(\frac{y_{3}^{2}}{4}, y_{3}\right), N\left(\frac{y_{4}^{2}}{4}, y_{4}\right)$, we get $\left\{\begin{array}{l}y_{3}+y_{4}=-4 k \\ y_{3} y_{4}=-4\end{array}\right.$
Therefore, $|M N|=\frac{y_{3}^{2}}{4}+\frac{y_{4}^{2}}{4}+2=4 k^{2}+4>6$.
(2) When points $A$, $B$, and $F$ are collinear,
$$
|A B|=\frac{y_{1}^{2}+y_{2}^{2}}{4}+2=6.
$$

Let the inclination angle of $A B$ be $\alpha$, then
$$
\frac{4}{\sin ^{2} \alpha}=6 \Rightarrow \sin ^{2} \alpha=\frac{2}{3},
$$

In this case, the common chord degenerates into a common tangent.
Thus, $M N \perp A B$,
$$
|M N|=\frac{4}{\sin ^{2}\left(\alpha+\frac{\pi}{2}\right)}=\frac{4}{\cos ^{2} \alpha}=12.
$$

In summary, $|M N| \in (6,12) \cup (12,+\infty)$.