$O P \perp O Q$.
Also, $C P \perp C Q$, so $P Q$ is the common hypotenuse of Rt $\triangle P O Q$ and $\mathrm{Rt} \triangle P C Q$. Therefore, the distance from the midpoint of line segment $P Q$ to points $O$ and $C$ is equal, meaning the x-coordinate of the midpoint of line segment $P Q$ is $\frac{1}{2}$.
Let the equation of line $P Q$ be $y=k x+b$.
Substituting into the ellipse equation gives
$$
\left(1+2 k^{2}\right) x^{2}+4 k b x+2 b^{2}-2=0 \text {. }
$$

Let $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$. Then
$$
x_{1}+x_{2}=-\frac{4 k b}{1+2 k^{2}}, x_{1} x_{2}=\frac{2 b^{2}-2}{1+2 k^{2}} \text {. }
$$

Since $x_{1}+x_{2}=1$, we have
$$
1+2 k^{2}=-4 k b \text {. }
$$

Thus, $y_{1} y_{2}=k^{2} x_{1} x_{2}+k b\left(x_{1}+x_{2}\right)+b^{2}$
$$
=\frac{2 k^{2} b^{2}-2 k^{2}}{1+2 k^{2}}+k b+b^{2} \text {. }
$$

Since $x_{1} x_{2}+y_{1} y_{2}=0$, we have
$$
\begin{array}{l}
\frac{2 b^{2}-2}{1+2 k^{2}}+\frac{2 k^{2} b^{2}-2 k^{2}}{1+2 k^{2}}+k b+b^{2}=0 \\
\Rightarrow 4 k^{2} b^{2}+2 k^{3} b-2 k^{2}+3 b^{2}+k b-2=0 .
\end{array}
$$

From equations (1) and (2), eliminating $b$ gives
$$
\begin{array}{l}
-20 k^{4}-20 k^{2}+3=0 \\
\Rightarrow k= \pm \sqrt{\frac{-5+2 \sqrt{10}}{10}} .
\end{array}
$$
10. (20 points) Given the ellipse $\Gamma: \frac{x^{2}}{2}+y^{2}=1$ with left and right foci $F_{1}$ and $F_{2}$, respectively, two perpendicular moving lines passing through the point $(1,0)$ intersect the ellipse $\Gamma$ at points $P$ and $Q$. When $\overrightarrow{P F_{1}}+\overrightarrow{P F_{2}}$ is perpendicular to $\overrightarrow{Q F_{1}}+\overrightarrow{Q F_{2}}$, find the slope of the line $P Q$.