Two non-intersecting circles, $\omega$ and $\Omega$, have centers $C_\omega$ and $C_\Omega$ respectively.  It is given that the radius of $\Omega$ is strictly larger than the radius of $\omega$.  The two common external tangents of $\Omega$ and $\omega$ intersect at a point $P$, and an internal tangent of the two circles intersects the common external tangents at $X$ and $Y$.  Suppose that the radius of $\omega$ is $4$, the circumradius of $\triangle PXY$ is $9$, and $XY$ bisects $\overline{PC_\Omega}$.  Compute $XY$.
1. **Understanding the Problem:**
   We are given two non-intersecting circles, $\omega$ and $\Omega$, with centers $C_\omega$ and $C_\Omega$ respectively. The radius of $\Omega$ is larger than the radius of $\omega$. The two common external tangents of $\Omega$ and $\omega$ intersect at a point $P$. An internal tangent of the two circles intersects the common external tangents at points $X$ and $Y$. We are given:
   - Radius of $\omega$ is $4$.
   - Circumradius of $\triangle PXY$ is $9$.
   - $XY$ bisects $\overline{PC_\Omega}$.

2. **Relating the Problem to a Triangle:**
   The problem can be translated into a geometric problem involving a triangle $ABC$ where:
   - $A$ is the point $P$.
   - $B$ and $C$ are points $X$ and $Y$ respectively.
   - The circumradius $R$ of $\triangle ABC$ is $9$.
   - The inradius $r$ of $\triangle ABC$ is $4$.
   - $XY$ bisects $\overline{PC_\Omega}$, implying that $XY$ is the $A$-altitude of $\triangle ABC$.

3. **Using Known Geometric Properties:**
   - The $A$-exradius $r_a$ of $\triangle ABC$ is equal to the $A$-altitude $h_a$.
   - From the property $r_a = h_a$, we derive that $2a = b + c$ where $a$, $b$, and $c$ are the sides of $\triangle ABC$ opposite to vertices $A$, $B$, and $C$ respectively.

4. **Using Euler's Theorem:**
   - Euler's theorem states that the distance $OI$ between the circumcenter $O$ and the incenter $I$ of a triangle is given by:
     \[
     OI = \sqrt{R(R - 2r)}
     \]
     Substituting $R = 9$ and $r = 4$:
     \[
     OI = \sqrt{9(9 - 8)} = \sqrt{9 \cdot 1} = 3
     \]

5. **Calculating $AI$:**
   - Using the Pythagorean theorem in $\triangle AOI$ where $O$ is the circumcenter, $I$ is the incenter, and $A$ is a vertex:
     \[
     AI = \sqrt{AO^2 - OI^2} = \sqrt{9^2 - 3^2} = \sqrt{81 - 9} = \sqrt{72} = 6\sqrt{2}
     \]

6. **Finding $\sin \left( \frac{A}{2} \right)$ and $\cos \left( \frac{A}{2} \right)$:**
   - Using the inradius $r$ and $AI$:
     \[
     \sin \left( \frac{A}{2} \right) = \frac{r}{AI} = \frac{4}{6\sqrt{2}} = \frac{2}{3\sqrt{2}}
     \]
     \[
     \cos \left( \frac{A}{2} \right) = \sqrt{1 - \sin^2 \left( \frac{A}{2} \right)} = \sqrt{1 - \left( \frac{2}{3\sqrt{2}} \right)^2} = \sqrt{1 - \frac{2}{9}} = \sqrt{\frac{7}{9}} = \frac{\sqrt{7}}{3}
     \]

7. **Calculating $\sin(A)$:**
   - Using the double-angle formula:
     \[
     \sin(A) = 2 \sin \left( \frac{A}{2} \right) \cos \left( \frac{A}{2} \right) = 2 \cdot \frac{2}{3\sqrt{2}} \cdot \frac{\sqrt{7}}{3} = \frac{4\sqrt{7}}{9\sqrt{2}} = \frac{4\sqrt{14}}{18} = \frac{2\sqrt{14}}{9}
     \]

8. **Using the Law of Sines:**
   - The side $a$ opposite to angle $A$ is:
     \[
     a = 2R \sin(A) = 2 \cdot 9 \cdot \frac{2\sqrt{14}}{9} = 4\sqrt{14}
     \]

9. **Conclusion:**
   - The length of $XY$ is $4\sqrt{14}$.

The final answer is $\boxed{4\sqrt{14}}$