11. (20 points) In the Cartesian coordinate system $x O y$, $A B$ is 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 $O$, $A$, and $B$). If $P F$ bisects $\angle A P B$, find all possible values of $|P F|$.

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11. Let \( A\left(\frac{y_{1}^{2}}{4}, y_{1}\right), B\left(\frac{y_{2}^{2}}{4}, y_{2}\right), P\left(\frac{y_{3}^{2}}{4}, y_{3}\right) \).

From the conditions, we know that \( y_{1}, y_{2}, y_{3} \) are pairwise distinct and non-zero. Let \( l_{A B}: x=t y+1 \), and by combining with the parabola equation, we get \( y^{2}-4 t y-4=0 \).
Thus, \( y_{1} y_{2}=-4 \).
Notice that, the circumcircle of \( \triangle A O B \) passes through point \( O \), so we can set the equation of the circle as \( x^{2}+y^{2}+d x+e y=0 \), and combining with \( x=\frac{y^{2}}{4} \) we get
\[
\frac{y^{4}}{16}+\left(1+\frac{d}{4}\right) y^{2}+e y=0 .
\]

This quartic equation has four distinct real roots \( y_{1}, y_{2}, y_{3}, 0 \).
Thus, by Vieta's formulas, we have
\[
y_{1}+y_{2}+y_{3}+0=0 \text{. }
\]

Therefore, \( y_{3}=-\left(y_{1}+y_{2}\right) \).
Since \( P F \) bisects \( \angle A P B \), by the Angle Bisector Theorem, we have
\[
\frac{|P A|}{|P B|}=\frac{|F A|}{|F B|}=\frac{\left|y_{1}\right|}{\left|y_{2}\right|} \text{. }
\]

Combining equations (1) and (2), we get
\[
\begin{aligned}
\frac{y_{1}^{2}}{y_{2}^{2}} & =\frac{|P A|^{2}}{|P B|^{2}}=\frac{\left(\frac{y_{3}^{2}}{4}-\frac{y_{1}^{2}}{4}\right)^{2}+\left(y_{3}-y_{1}\right)^{2}}{\left(\frac{y_{3}^{2}}{4}-\frac{y_{2}^{2}}{4}\right)^{2}+\left(y_{3}-y_{2}\right)^{2}} \\
& =\frac{\left(\left(y_{1}+y_{2}\right)^{2}-y_{1}^{2}\right)^{2}+16\left(2 y_{1}+y_{2}\right)^{2}}{\left(\left(y_{1}+y_{2}\right)^{2}-y_{2}^{2}\right)^{2}+16\left(2 y_{2}+y_{1}\right)^{2}} \\
& =\frac{\left(y_{2}^{2}-8\right)^{2}+16\left(4 y_{1}^{2}+y_{2}^{2}-16\right)}{\left(y_{1}^{2}-8\right)^{2}+16\left(4 y_{2}^{2}+y_{1}^{2}-16\right)} \\
& =\frac{y_{2}^{4}+64 y_{1}^{2}-192}{y_{1}^{4}+64 y_{2}^{2}-192},
\end{aligned}
\]

Thus,
\[
\begin{array}{l}
y_{1}^{6}+64 y_{1}^{2} y_{2}^{2}-192 y_{1}^{2}=y_{2}^{6}+64 y_{2}^{2} y_{1}^{2}-192 y_{2}^{2} \\
\text{ Hence } \left(y_{1}^{2}-y_{2}^{2}\right)\left(y_{1}^{4}+y_{1}^{2} y_{2}^{2}+y_{2}^{4}-192\right)=0 .
\end{array}
\]

When \( y_{1}^{2}=y_{2}^{2} \), \( y_{2}=-y_{1} \), so \( y_{3}=0 \), in this case, point \( P \) coincides with \( O \), which contradicts the conditions.
When \( y_{1}^{4}+y_{1}^{2} y_{2}^{2}+y_{2}^{4}-192=0 \), combining with equation (1) we have \( \left(y_{1}^{2}+y_{2}^{2}\right)^{2}=192+\left(y_{1} y_{2}\right)^{2}=208 \).
Also, \( y_{1}^{2}+y_{2}^{2}=4 \sqrt{13}>8=\left|2 y_{1} y_{2}\right| \), so there exist real numbers \( y_{1}, y_{2} \) satisfying \( y_{1}^{2}+y_{2}^{2}=4 \sqrt{13} \) and the corresponding points \( A, B \). At this time, combining equations (1) and (2), we have
\[
\begin{array}{l}
|P F|=\frac{y_{3}^{2}}{4}+1=\frac{\left(y_{1}+y_{2}\right)^{2}+4}{4} \\
=\frac{y_{1}^{2}+y_{2}^{2}-4}{4}=\frac{\sqrt{208}-4}{4}=\sqrt{13}-1 .
\end{array}
\]