Point $A$ lies on the circumference of a circle $\Omega$ with radius $78$.  Point $B$ is placed such that $AB$ is tangent to the circle and $AB=65$, while point $C$ is located on $\Omega$ such that $BC=25$.  Compute the length of $\overline{AC}$.
1. **Identify the given information and draw the diagram:**
   - Circle $\Omega$ with radius $78$.
   - Point $A$ on the circumference of $\Omega$.
   - Point $B$ such that $AB$ is tangent to $\Omega$ and $AB = 65$.
   - Point $C$ on $\Omega$ such that $BC = 25$.

2. **Use the Power of a Point theorem:**
   - Extend $BC$ to hit the circle $\Omega$ at another point $P$.
   - By the Power of a Point theorem, we have:
     \[
     AB^2 = BC \times BP
     \]
   - Substituting the given values:
     \[
     65^2 = 25 \times BP
     \]
   - Solving for $BP$:
     \[
     4225 = 25 \times BP \implies BP = \frac{4225}{25} = 169
     \]

3. **Verify the relationship between $A$, $B$, and $P$:**
   - Since $AB$ is tangent to the circle at $A$, and $BP = 169$, we check if $P$ is the antipode of $A$.
   - The distance from $A$ to the center of the circle is the radius, $78$.
   - The distance from $B$ to the center of the circle is:
     \[
     \sqrt{78^2 + 65^2} = \sqrt{6084 + 4225} = \sqrt{10309} = 101.54
     \]
   - Since $BP = 169$, and $169 = 2 \times 78 + 13$, $P$ is indeed the antipode of $A$.

4. **Determine the length of $CP$:**
   - Since $P$ is the antipode of $A$, $AP = 2 \times 78 = 156$.
   - Given $BP = 169$ and $BC = 25$, we find $CP$:
     \[
     CP = BP - BC = 169 - 25 = 144
     \]

5. **Use the Pythagorean Theorem to find $AC$:**
   - Since $\angle ACP = 90^\circ$ (as $P$ is the antipode of $A$), we use the Pythagorean Theorem:
     \[
     AC^2 + CP^2 = AP^2
     \]
   - Substituting the known values:
     \[
     AC^2 + 144^2 = 156^2
     \]
     \[
     AC^2 + 20736 = 24336
     \]
     \[
     AC^2 = 3600
     \]
     \[
     AC = \sqrt{3600} = 60
     \]

The final answer is $\boxed{60}$.