11. In the Cartesian coordinate system $x O y$, let $A B$ be a chord of the parabola $y^{2}=4 x$ passing through the point $F(1,0)$, and the circumcircle of $\triangle A O B$ intersects the parabola at point $P$ (different from points $O, A, B$). If $P F$ bisects $\angle A P B$, find all possible values of $|P F|$.
Given $A\left(\frac{a^{2}}{4}, a\right), B\left(\frac{b^{2}}{4}, b\right), P\left(\frac{c^{2}}{4}, c\right)$, and the circumcircle equation of $\triangle A O B$ is $x^{2}+y^{2}+d x+e y=0$.
Substituting $x=\frac{y^{2}}{4}$, we get $\frac{y^{4}}{16}+\left(1+\frac{d}{4}\right) y^{2}+e y=0$.
Since $a, b, c, 0$ are the four roots of the above equation,
then $a+b+c+0=0 \Rightarrow c=-(a+b)$.
Noting that the line $A B$ passes through the focus $F(1,0)$ of the parabola $y^{2}=4 x$, let $A B: x=t y+1$,
Combining $\left\{\begin{array}{l}y^{2}=4 x, \\ x=t y+1\end{array} \Rightarrow y^{2}-4 t y-4=0 \Rightarrow a b=-4\right.$.
$P F$ bisects $\angle A P B \Rightarrow \frac{|P A|}{|P B|}=\frac{|F A|}{|F B|}=-\frac{a}{b} \Rightarrow \frac{\left(\frac{c^{2}}{4}-\frac{a^{2}}{4}\right)^{2}+(c-a)^{2}}{\left(\frac{c^{2}}{4}-\frac{b^{2}}{4}\right)^{2}+(c-b)^{2}}=\frac{a^{2}}{b^{2}}$ $\Rightarrow \frac{(c-a)^{2}\left[(c+a)^{2}+16\right]}{(c-b)^{2}\left[(c+b)^{2}+16\right]}=\frac{(2 a+b)^{2}\left(b^{2}+16\right)}{(a+2 b)^{2}\left(a^{2}+16\right)}=\frac{a^{2}}{b^{2}}$ $\Rightarrow b^{2}\left(b^{2}+16\right)(2 a+b)^{2}=a^{2}\left(a^{2}+16\right)(a+2 b)^{2}$ $\Rightarrow a^{6}-b^{6}+(16+4 a b)\left(a^{4}-b^{4}\right)+\left(4 a^{2} b^{2}+64 a b\right)\left(a^{2}-b^{2}\right)=0$.
If $a^{2}=b^{2} \Rightarrow a+b=0 \Rightarrow c=0$, in this case point $P$ coincides with point $O$, which is a contradiction.
Thus, $a^{4}+a^{2} b^{2}+b^{4}+(16+4 a b)\left(a^{2}+b^{2}\right)+4 a^{2} b^{2}+64 a b=0$
$\Rightarrow\left(a^{2}+b^{2}\right)^{2}+(16+4 a b)\left(a^{2}+b^{2}\right)+3 a^{2} b^{2}+64 a b=0$
$\Rightarrow\left(a^{2}+b^{2}\right)^{2}+3 \times 16-64 \times 4=\left(a^{2}+b^{2}\right)^{2}-16 \times 13=0$
$\Rightarrow a^{2}+b^{2}=4 \sqrt{13}$. Therefore, $a^{2}, b^{2}$ are the two roots of the equation $x^{2}-4 \sqrt{13} x+16=0$,
Solving, we get $a^{2}=2 \sqrt{13}+6, b^{2}=2 \sqrt{13}-6$.
Thus, $|P F|=\frac{c^{2}}{4}+1=\frac{(a+b)^{2}}{4}+1=\frac{a^{2}+b^{2}+2 a b+4}{4}=\frac{4 \sqrt{13}-4}{4}=\sqrt{13}-1$.