11. (This question is worth 20 points) As shown in the figure, 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 $C$ 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 to $C$, and $|P Q|=2$. Circles $C_{1}$ and $C_{2}$ are both tangent to 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.
Let the equation of the parabola $C$ be $y^{2}=2 p x (p>0)$, the coordinates of point $Q$ be $(-a, 0) (a>0)$, and the centers of $C_{1}, C_{2}$ be $O_{1}(x_{1}, y_{1}), O_{2}(x_{2}, y_{2})$ respectively.
Assume the equation of the line $P Q$ is $x=m y-a (m>0)$. By combining this with the equation of $C$ and eliminating $x$, we get $y^{2}-2 p m y+2 p a=0$.

Since $P Q$ is tangent to $C$ at point $P$, the discriminant of the above equation is $\Delta=4 p^{2} m^{2}-4 \cdot 2 p a=0$, solving which gives $m=\sqrt{\frac{2 a}{p}}$. Therefore, the coordinates of point $P$ are $\left(x_{P}, y_{P}\right)=(a, \sqrt{2 p a})$. Thus,
$$
|P Q|=\sqrt{1+m^{2}} \cdot\left|y_{P}-0\right|=\sqrt{1+\frac{2 a}{p}} \cdot \sqrt{2 p a}=\sqrt{2 a(p+2 a)} .
$$

From $|P Q|=2$, we have
$$
4 a^{2}+2 p a=4 .
$$

Noting that $O P$ is tangent to the circles $C_{1}, C_{2}$ at point $P$, we have $O P \perp O_{1} O_{2}$. Let the circles $C_{1}, C_{2}$ be tangent to the $x$-axis at points $M, N$ respectively, then $O O_{1}, O O_{2}$ are the angle bisectors of $\angle P O M, \angle P O N$, hence $\angle O_{1} O O_{2}=90^{\circ}$. Therefore, by the projection theorem,
$$
\begin{aligned}
y_{1} y_{2} & =O_{1} M \cdot O_{2} N=O_{1} P \cdot O_{2} P=O P^{2} \\
& =x_{P}^{2}+y_{P}^{2}=a^{2}+2 p a .
\end{aligned}
$$

Combining (1), we get
$$
y_{1} y_{2}=a^{2}+2 p a=4-3 a^{2} .
$$
10 points
Since $O_{1}, P, O_{2}$ are collinear, we have
$$
\frac{y_{1}-\sqrt{2 p a}}{\sqrt{2 p a}-y_{2}}=\frac{y_{1}-y_{P}}{y_{P}-y_{2}}=\frac{O_{1} P}{P O_{2}}=\frac{O_{1} M}{O_{2} N}=\frac{y_{1}}{y_{2}},
$$

Simplifying, we get
$$
y_{1}+y_{2}=\frac{2}{\sqrt{2 p a}} \cdot y_{1} y_{2} .
$$
15 points
Let $T=y_{1}^{2}+y_{2}^{2}$, then the sum of the areas of circles $C_{1}, C_{2}$ is $\pi T$. According to the problem, we only need to consider the case where $T$ is minimized.
From (2) and (3), 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 $t=1-a^{2}$. Since $4 t=4-4 a^{2}=2 p a>0$, we have $t>0$. Thus,
$$
T=\frac{(3 t+1)(t+1)}{t}=3 t+\frac{1}{t}+4 \geq 2 \sqrt{3 t \cdot \frac{1}{t}}+4=2 \sqrt{3}+4
$$

The equality holds if and only if $t=\frac{\sqrt{3}}{3}$, at which point $a=\sqrt{1-t}=\sqrt{1-\frac{1}{\sqrt{3}}}$. Therefore, combining (1), we get
$$
\frac{p}{2}=\frac{1-a^{2}}{a}=\frac{t}{\sqrt{1-\frac{1}{\sqrt{3}}}}=\frac{\sqrt{3} t}{\sqrt{3-\sqrt{3}}}=\frac{1}{\sqrt{3-\sqrt{3}}},
$$

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