4. As shown in Figure 4, given that the radii of circles $\odot O_{1}$ and $\odot O_{2}$, which intersect at points $A$ and $B$, are $5$ and $7$ respectively, and $O_{1} O_{2}=6$. A line through point $A$ intersects the two circles at points $C$ and $D$. $P$ and $O$ are the midpoints of line segments $CD$ and $O_{1} O_{2}$, respectively. Then the length of $OP$ is
$4.2 \sqrt{7}$.
As shown in Figure 9, draw the diameter $A E$ of $\odot O_{1}$ through point $A$, connect $E B$ and extend it to intersect $\odot O_{2}$ at point $F$, connect $A O$ and extend it to intersect $E F$ at point $G$, connect $A B$, $C E$, $D F$, and $P G$. Let $A B$ intersect $O_{1} O_{2}$ at point $H$. Since $A E$ is the diameter of $\odot O_{1}$
$$
\Rightarrow \angle A B E=90^{\circ} \Rightarrow \angle A B F=90^{\circ}
$$
$\Rightarrow A F$ is the diameter of $\odot O_{2}$, $O_{1} O_{2} / / E F$
$$
\Rightarrow \frac{O_{1} O}{E G}=\frac{A O}{A G}=\frac{O O_{2}}{G F} \text {. }
$$

Since $O$ is the midpoint of $O_{1} O_{2}$, we have $O_{1} O=0 O_{2}=3$.
Thus, $G E=G F$, meaning $G$ is the midpoint of $E F$.
Since $A E$ and $A F$ are the diameters of $\odot O_{1}$ and $\odot O_{2}$ respectively, then $\angle C=\angle D=90^{\circ}, C E / / D F$.
Since $P$ and $G$ are the midpoints of $C D$ and $E F$ respectively
$$
\Rightarrow P G / / C E / / D F \Rightarrow \angle A P G=\angle C=90^{\circ} \text {. }
$$

Since $O_{1} O / / E G \Rightarrow \frac{A O}{A G}=\frac{A O_{1}}{A E}=\frac{1}{2} \Rightarrow A O=\frac{1}{2} A G$ $\Rightarrow O$ is the midpoint of $A G$ $\Rightarrow O P=\frac{1}{2} A G=A O$.
Let $O_{1} H=x, A H=y$, then $O_{2} H=6-x$.
In $\mathrm{Rt} \triangle A O_{1} H$ and $\mathrm{Rt} \triangle A O_{2} H$, by the Pythagorean theorem we get
$$
\left\{\begin{array}{l}
O_{1} H^{2}+A H^{2}=O_{1} A^{2}, \\
O_{2} H^{2}+A H^{2}=O_{2} A^{2},
\end{array}\right.
$$

which is $\left\{\begin{array}{l}x^{2}+y^{2}=5^{2}, \\ (6-x)^{2}+y^{2}=7^{2}\end{array}\right.$
Solving these equations, we get $x=1, y=2 \sqrt{6}$.
Therefore, $O H=O_{1} O-O_{1} H=3-1=2$.
Thus, $O A=\sqrt{O H^{2}+A H^{2}}=\sqrt{2^{2}+(2 \sqrt{6})^{2}}=2 \sqrt{7}$.
Hence, $O P=2 \sqrt{7}$.