Four. (20 points) 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 $l$, and $\odot Q$ 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 $\angle MAN$.

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Translation:

Four. (20 points) 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 $l$, and $\odot Q$ 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 $\angle MAN$.
$$
\begin{array}{rl} 
& \text{Let } l \text{ be the } x \text{-axis, and the perpendicular from point } P \text{ to } l \text{ be the } y \text{-axis, as shown in Figure 6.} \\
& \text{Suppose the coordinates of } Q \text{ are } (x, 0), \text{ and point } A \text{ is at } (k, h). \\
& \text{The radius of } \odot Q \text{ is } r. \text{ Then,} \\
& M(x-r, 0), \\
N(x+r, 0), P(0,2) . \\
P Q & =\sqrt{x^{2}+2^{2}} \\
& =1+r . \\
\therefore x & = \pm \sqrt{r^{2}+2 r-3} . \\
\therefore \tan \angle M A N=\frac{k_{A N}-k_{A M}}{1+k_{A N} k_{A M}} \\
= & \frac{0-h}{1+\frac{0-h}{x+r-k} \cdot \frac{0-h}{x-r-k}}=\frac{2 r h}{(x-k)^{2}-r^{2}+h^{2}} \\
= & \frac{0-h}{h^{2}+k^{2}-3+2 r \mp 2 k \sqrt{r^{2}+2 r-3}} . \\
\quad \text { Let } & 2 m=h^{2}+k^{2}-3, \tan \angle M A N=\frac{1}{n} . \\
\therefore m+r \mp k \sqrt{r^{2}+2 r-3}=n h r, \\
m & m(1-n h) r= \pm k \sqrt{r^{2}+2 r-3} . \\
&
\end{array}
$$

Square both sides, we get
$$
\begin{array}{c}
m^{2}+2 m(1-n h) r+(1-n h)^{2} r^{2} \\
=k^{2} r^{2}+2 k^{2} r-3 k^{2} .
\end{array}
$$
Since for any real number $r \geqslant 1$, the above equation always holds,
$$
\therefore\left\{\begin{array}{l}
m^{2}=-3 k^{2}, \\
2 m(1-n h)=2 k^{2}, \\
(1-n h)^{2}=k^{2} .
\end{array}\right.
$$

From equation (1), we get $m=0, k=0$.
From equation (3), we get $n=\frac{1}{h}$.
From $2 m=h^{2}+k^{2}-3$, we get $h= \pm \sqrt{3}$.
$$
\therefore \tan \angle M A N=\frac{1}{n}=h= \pm \sqrt{3} \text {. }
$$
$\therefore \angle M A N=60^{\circ}$ or $120^{\circ}$ (discard $120^{\circ}$ when $Q(0,0), r=1$, $\angle M A N=60^{\circ}$).
Hence, $\angle M A N=60^{\circ}$.