II. (20 points) Let a line $l$ that does not pass through the origin $O$ intersect the circle $x^{2}+y^{2}=1$ at two distinct points $P$ and $Q$. If the slope of line $PQ$ is the geometric mean of the slopes of lines $OP$ and $OQ$, find the range of the area $S$ of $\triangle POQ$.
Let $l_{P Q}: y=k x+b(k \neq 0, b \neq 0)$, substituting into $x^{2}+y^{2}=1$, we get
$$
\left(k^{2}+1\right) x^{2}+2 k b x+b^{2}-1=0 .
$$

From $\Delta=4 k^{2} b^{2}-4\left(k^{2}+1\right)\left(b^{2}-1\right)>0$
$$
\Rightarrow b^{2}<k^{2}+1 \text {. }
$$

Let $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)\left(x_{1} x_{2} \neq 0\right)$. Then
$$
x_{1}+x_{2}=-\frac{2 k b}{k^{2}+1}, x_{1} x_{2}=\frac{b^{2}-1}{k^{2}+1} \text {. }
$$

Thus, $k_{O P} k_{O Q}=\frac{y_{1}}{x_{1}} \cdot \frac{y_{2}}{x_{2}}=\frac{\left(k x_{1}+b\right)\left(k x_{2}+b\right)}{x_{1} x_{2}}$
$$
=k^{2}+\frac{k b\left(x_{1}+x_{2}\right)+b^{2}}{x_{1} x_{2}} \text {. }
$$

Since $k_{O P} k_{O Q}=k^{2}$, we have
$$
\begin{array}{l}
k b\left(x_{1}+x_{2}\right)+b^{2}=0 \\
\Rightarrow x_{1}+x_{2}=-\frac{b}{k}=-\frac{2 k b}{k^{2}+1} .
\end{array}
$$

Solving, we get $k= \pm 1$.
The distance from the origin $O$ to the line $P Q$ is
$$
d=\frac{|b|}{\sqrt{k^{2}+1}}=\frac{|b|}{\sqrt{2}},
$$

Thus, $|P Q|=2 \sqrt{1-d^{2}}=2 \sqrt{1-\frac{b^{2}}{2}}$.
Therefore, $S=\frac{1}{2} d|P Q|=\frac{|b|}{\sqrt{2}} \sqrt{1-\frac{b^{2}}{2}}$
$$
\begin{array}{l}
=\sqrt{\frac{b^{2}}{2}\left(1-\frac{b^{2}}{2}\right)} \\
\leqslant \frac{\frac{b^{2}}{2}+\left(1-\frac{b^{2}}{2}\right)}{2}=\frac{1}{2} .
\end{array}
$$

Since $x_{1} x_{2}=\frac{b^{2}-1}{k^{2}+1} \neq 0$, we have $b \neq 1$.
Therefore, the equality in (1) does not hold.
Hence, the range of the area $S$ of $\triangle P O Q$ is $\left(0, \frac{1}{2}\right)$.