11. (20 points) As shown in Figure 1, in the Cartesian coordinate system $x O y$, $F$ is a moving point on the positive half-axis of the $x$-axis. A parabola $C$ is constructed with $F$ as the focus and $O$ as the vertex. Let $P$ be a point on the parabola $C$ in the first quadrant, and $Q$ be a point on the negative half-axis of the $x$-axis such that $P Q$ is a tangent line to the parabola $C$, and $|P Q|=2$. Circles $C_{1}$ and $C_{2}$ both touch the line $O P$ at point $P$, and both are tangent to the $x$-axis. Find the coordinates of point $F$ such that the sum of the areas of circles $C_{1}$ and $C_{2}$ is minimized.
11. Let the parabola $C: y^{2}=2 p x(p>0)$, point $Q(-a, 0)(a>0)$, and the centers of circles $C_{1}$ and $C_{2}$ be $O_{1}\left(x_{1}, y_{1}\right)$ and $O_{2}\left(x_{2}, y_{2}\right)$, respectively.
Let $l_{P Q}: x=m y-a(m>0)$, and combine it with the equation of the parabola $C$, eliminating $x$ we get
$$
y^{2}-2 p m y+2 p a=0 \text {. }
$$

Since $P Q$ is tangent to the parabola $C$ at point $P$, the discriminant of equation (1) is
$$
\Delta=4 p^{2} m^{2}-4 \times 2 p a=0 \Rightarrow m=\sqrt{\frac{2 a}{p}} \text {. }
$$

Thus, the point $P\left(x_{P}, y_{P}\right)=(a, \sqrt{2 p a})$.
Hence, $|P Q|=\sqrt{1+m^{2}}\left|y_{P}-0\right|$
$$
=\sqrt{1+\frac{2 a}{p}} \sqrt{2 p a}=\sqrt{2 a(p+2 a)} \text {. }
$$

Given $|P Q|=2$
$$
\Rightarrow 4 a^{2}+2 p a=4 \text {. }
$$

Note that $O P$ is tangent to circles $C_{1}$ and $C_{2}$ at point $P$.
Thus, $O P \perp O_{1} O_{2}$.
Assume circles $C_{1}$ and $C_{2}$ are tangent to the $x$-axis at points $M$ and $N$, respectively, as shown in Figure 5.

Then $O O_{1}$ and $O O_{2}$ are the angle bisectors of $\angle P O M$ and $\angle P O N$, respectively. Hence, $\angle O_{1} \mathrm{OO}_{2}=90^{\circ}$.
By the projection theorem, we have
$$
\begin{array}{c}
y_{1} y_{2}=O_{1} M \cdot O_{2} N=O_{1} P \cdot O_{2} P \\
=O P^{2}=x_{P}^{2}+y_{P}^{2}=a^{2}+2 p a .
\end{array}
$$

Combining equation (2), we get
$$
y_{1} y_{2}=a^{2}+2 p a=4-3 a^{2} \text {. }
$$

Since $O_{1}$, $P$, and $O_{2}$ are collinear, we have
$$
\begin{array}{l}
\frac{y_{1}-\sqrt{2 p a}}{\sqrt{2 p a}-y_{2}}=\frac{y_{1}-y_{P}}{y_{P}-y_{2}}=\frac{O_{1} P}{P O_{2}}=\frac{O_{1} M}{O_{2} N}=\frac{y_{1}}{y_{2}} \\
\Rightarrow y_{1}+y_{2}=\frac{2}{\sqrt{2 p a}} y_{1} y_{2} .
\end{array}
$$

Let $T=y_{1}^{2}+y_{2}^{2}$, then the sum of the areas of circles $C_{1}$ and $C_{2}$ is $\pi T$.
According to the problem, we only need to consider the case where $T$ is minimized.
From equations (3) and (4), we have
$$
\begin{aligned}
T & =\left(y_{1}+y_{2}\right)^{2}-2 y_{1} y_{2}=\frac{4}{2 p a} y_{1}^{2} y_{2}^{2}-2 y_{1} y_{2} \\
& =\frac{4}{4-4 a^{2}}\left(4-3 a^{2}\right)^{2}-2\left(4-3 a^{2}\right) \\
& =\frac{\left(4-3 a^{2}\right)\left(2-a^{2}\right)}{1-a^{2}} .
\end{aligned}
$$

Let $t=1-a^{2}$.
By $4 t=4-4 a^{2}=2 p a>0 \Rightarrow t>0$
$\Rightarrow T=\frac{(3 t+1)(t+1)}{t}=3 t+\frac{1}{t}+4$
$$
\geqslant 2 \sqrt{3 t \cdot \frac{1}{t}}+4=2 \sqrt{3}+4 \text {. }
$$

Equality holds if and only if $t=\frac{\sqrt{3}}{3}$.
At this time, $a=\sqrt{1-t}=\sqrt{1-\frac{1}{\sqrt{3}}}$.
Combining equation (2), we get
$$
\begin{array}{l}
\frac{p}{2}=\frac{1-a^{2}}{a}=\frac{t}{\sqrt{1-\frac{1}{\sqrt{3}}}} \\
=\frac{\sqrt{3} t}{\sqrt{3-\sqrt{3}}}=\frac{1}{\sqrt{3-\sqrt{3}}} .
\end{array}
$$

Thus, the point $F\left(\frac{p}{2}, 0\right)=\left(\frac{1}{\sqrt{3-\sqrt{3}}}, 0\right)$.