10. Given a fixed circle $\odot P$ with radius 1, the distance from the center $P$ to a fixed line $l$ is 2, $Q$ is a moving point on $l$, $\odot Q$ is externally tangent to $\odot P$, $\odot Q$ intersects $l$ at points $M$ and $N$, for any diameter $MN$, there is a fixed point $A$ in the plane such that $\angle MAN$ is a constant, find the degree measure of $\angle MAN$.
10. Solution: Taking $l$ as the $x$-axis and the perpendicular from point $P$ to $l$ as the $y$-axis, we establish a rectangular coordinate system as shown in the figure. Let the coordinates of $Q$ be $(x, 0)$, the coordinates of point $A$ be $(k, h)$, and the radius of $\odot Q$ be $r$. Then: $M(x-r, 0)$, $N(x+r, 0)$, $P(2,0)$, and $PQ=\sqrt{x^{2}+2^{2}}=1+r$. Therefore, $x= \pm \sqrt{r^{2}+2 r-3}$,
$$
\begin{array}{c}
\therefore \tan \angle M A N=\frac{k_{A N}-k_{A M}}{1+k_{A N} \cdot k_{A M}}=\frac{\frac{0-h}{x+r-k}-\frac{0-h}{x-r-k}}{1+\frac{0-h}{x+r-k} \cdot \frac{0-h}{x-r-k}} \\
=\frac{2 r h}{(x-k)^{2}-r^{2}+h^{2}}=\frac{2 r h}{\left( \pm \sqrt{r^{2}+2 r-3}-k\right)^{2}-r^{2}+h^{2}} \\
=\frac{2 r h}{h^{2}+k^{2}-3+2 r \mp 2 k \sqrt{r^{2}+2 r-3}} \text {, let } 2 m=h^{2}+k^{2}-3, \tan \angle M A N=\frac{1}{n}, \text { so }
\end{array}
$$

with $m+r \mp k \sqrt{r^{2}+2 r-3}=n h r, \therefore m+(1-n h) r= \pm k \sqrt{r^{2}+2 r-3}$, squaring both sides, we get: $m^{2}+2 m(1-n h) r$ $+(1-n h)^{2} r^{2}=k^{2} r^{2}+2 k^{2} r-3 k^{2}$, since for any real number $r \geqslant 1$, the above equation always holds, we have $\left\{\begin{array}{l}m^{2}=-3 k^{2}(1) \\ 2 m(1-n h)=2 k^{2}(2) \text {, } \\ (1-n h)^{2}=k^{2}(3)\end{array}\right.$
From (1) and (2), we get $m=0, k=0$, from (3), we get $n=\frac{1}{h}$. From $2 m=h^{2}+k^{2}-3$ we get $h= \pm \sqrt{3}$, so $\tan \angle M A N=\frac{1}{n}$ $=h= \pm \sqrt{3}$. Therefore, $\angle M A N=60^{\circ}$ or $120^{\circ}$ (discard the latter) (when $Q(0,0), r=1$, $\angle M A N=60^{\circ}$), hence $\angle M A N=60^{\circ}$.