11. 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 $F\left(\frac{p}{2}, 0\right)$, then the parabola equation is $y^{2}=2 p x, P\left(2 p t^{2}, 2 p t\right) \Rightarrow P Q: 2 t y=x+2 p t^{2}$. Let $y=0 \Rightarrow Q\left(-2 p t^{2}, 0\right)$. Thus, $|P Q|^{2}=16 p^{2} t^{4}+4 p^{2} t^{2}=4 \Rightarrow 4 p^{2} t^{4}+p^{2} t^{2}=1$.
Since the distances from $C_{1}, C_{2}$ to point $P$ are equal to their distances to the $x$-axis, $C_{1}, C_{2}$ lie on the parabola $C^{\prime}$ with focus at $P$ and the $x$-axis as the directrix, and the corresponding $p^{\prime}=2 p t$.
By the problem, $O P \perp C_{1} C_{2} \Rightarrow C_{1} C_{2}$ makes an angle with the $y$-axis equal to the inclination angle $\theta$ of line $O P$, so $\tan \theta=k_{O P}=\frac{2 p t}{2 p t^{2}}=\frac{1}{t} \Rightarrow \cot \theta=t$.
For the parabola $C^{\prime}$, we have $\left|P C_{1}\right|=r_{1}=\frac{2 p t}{1-\cos \theta},\left|P C_{2}\right|=r_{2}=\frac{2 p t}{1+\cos \theta}$.
Thus, $r_{1}^{2}+r_{2}^{2}=\left(\frac{2 p t}{1-\cos \theta}\right)^{2}+\left(\frac{2 p t}{1+\cos \theta}\right)^{2}=\frac{8 p^{2} t^{2}\left(1+\cos ^{2} \theta\right)}{\left(1-\cos ^{2} \theta\right)^{2}}$
$=\frac{8 p^{2} t^{2}\left(2 \cos ^{2} \theta+\sin ^{2} \theta\right)\left(\cos ^{2} \theta+\sin ^{2} \theta\right)}{\sin ^{4} \theta}=8 p^{2} t^{2}\left(2 \cot ^{2} \theta+1\right)\left(\cot ^{2} \theta+1\right)$
$=8 p^{2} t^{2}\left(2 t^{2}+1\right)\left(t^{2}+1\right)=\left(16 p^{2} t^{4}+8 p^{2} t^{2}\right)\left(t^{2}+1\right)=\left(4 p^{2} t^{2}+4\right)\left(t^{2}+1\right)$
$=4 p^{2} t^{4}+4 p^{2} t^{2}+4 t^{2}+4=3 p^{2} t^{2}+4 t^{2}+5=\frac{3}{4 t^{2}+1}+4 t^{2}+5$
$=\frac{3}{4 t^{2}+1}+4 t^{2}+1+4 \geqslant 2 \sqrt{3}+4$, equality holds when $4 t^{2}+1=\sqrt{3}$.
Therefore, when the sum of the areas of circles $C_{1}$ and $C_{2}$ is minimized,
$$
p^{2} t^{2}=\frac{1}{4 t^{2}+1}=\frac{1}{\sqrt{3}} \Rightarrow p^{2}=\frac{1}{\sqrt{3} t^{2}}=\frac{4}{\sqrt{3}(\sqrt{3}-1)}=\frac{4}{3-\sqrt{3}} \Rightarrow \frac{p}{2}=\frac{1}{\sqrt{3-\sqrt{3}}} \text {. }
$$

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