Three. (20 points) Through the focus $\boldsymbol{F}(c, 0)$ of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, draw the line $y=(x-c) \tan \theta\left(\frac{\pi}{2}<\theta<\pi\right)$ intersecting the two directrices of the hyperbola at points $A$ and $B$. If the circle with $A B$ as its diameter passes through the origin, find $\theta$.
Solution 1: The directrices of the hyperbola are $x= \pm \frac{a^{2}}{c}$.
Substituting into the line equation, we get
$$
\begin{array}{l}
A\left(\frac{a^{2}}{c},-\frac{c^{2}-a^{2}}{c} \tan \theta\right), \\
B\left(-\frac{a^{2}}{c},-\frac{c^{2}+a^{2}}{c} \tan \theta\right) .
\end{array}
$$

Since the circle with diameter $AB$ passes through the origin, $\angle A O B=90^{\circ}$, so,
$$
\begin{array}{l}
0=\overrightarrow{O A} \cdot \overrightarrow{O B}=-\frac{a^{4}}{c^{2}}+\frac{c^{4}-a^{4}}{c^{2}} \tan ^{2} \theta \\
\Leftrightarrow a^{4}\left(1+\tan ^{2} \theta\right)=c^{4} \tan ^{2} \theta .
\end{array}
$$

Therefore, $\sin ^{2} \theta=\frac{a^{4}}{c^{4}}$.
Given $\frac{\pi}{2}<\theta<\pi$, we have $\theta=\pi-\arcsin \frac{a^{2}}{c^{2}}$.
Solution 2: The directrices of the hyperbola are $x= \pm \frac{a^{2}}{c}$, and the intersection points of the two directrices with the $x$-axis are $P\left(-\frac{a^{2}}{c}, 0\right), Q\left(\frac{a^{2}}{c}, 0\right)$, so $P Q=\frac{2 a^{2}}{c}$.
Given $\frac{\pi}{2}<\theta<\pi$, we have
$$
A B=\frac{P Q}{\cos (\pi-\theta)}=-\frac{2 a^{2}}{c \cos \theta} .
$$

The intersection point of line $l$ with the $y$-axis is $M(0,-c \tan \theta)$.
Since the circle with diameter $AB$ passes through the origin, $\angle A O B=90^{\circ}$, so, in the right triangle $\triangle A O B$, the median to the hypotenuse is half the hypotenuse, giving $O M=\frac{1}{2} A B$, i.e.,
$$
-c \tan \theta=\frac{1}{2}\left(-\frac{2 a^{2}}{c \cos \theta}\right) \text {. }
$$

Thus, $\sin \theta=\frac{a^{2}}{c^{2}}$.
Given $\frac{\pi}{2}<\theta<\pi$, we have $\theta=\pi-\arcsin \frac{a^{2}}{c^{2}}$.