Let $O_1$, $O_2$ be two circles with radii $R_1$ and $R_2$, and suppose the circles meet at $A$ and $D$. Draw a line $L$ through $D$ intersecting $O_1$, $O_2$ at $B$ and $C$. Now allow the distance between the centers as well as the choice of $L$ to vary. Find the length of $AD$ when the area of $ABC$ is maximized.
1. **Identify the given elements and setup the problem:**
   - Let \( O_1 \) and \( O_2 \) be the centers of two circles with radii \( R_1 \) and \( R_2 \) respectively.
   - The circles intersect at points \( A \) and \( D \).
   - A line \( L \) through \( D \) intersects \( O_1 \) at \( B \) and \( O_2 \) at \( C \).

2. **Define the angle and similarity:**
   - Let \( \angle ADB = \theta \).
   - Note that \( \angle AO_1B = \angle AO_2C = 2\theta \), so \( \triangle AO_1B \sim \triangle AO_2C \) since both are isosceles triangles.
   - By spiral similarity, \( \triangle ABC \sim \triangle AO_1O_2 \).

3. **Express the area of \( \triangle ABC \):**
   - The area of \( \triangle ABC \) can be written as:
     \[
     [ABC] = [AO_1O_2] \left( \frac{AB}{AO_1} \right)^2
     \]
   - Using the Extended Law of Sines, we have:
     \[
     [AO_1O_2] = \frac{1}{2} R_1 R_2 \sin \angle O_1AO_2
     \]
   - Therefore:
     \[
     [ABC] = \left( \frac{1}{2} R_1 R_2 \sin \angle O_1AO_2 \right) \left( \frac{AB}{R_1} \right)^2 = 2 R_1 R_2 \sin \angle O_1AO_2 \sin^2 \theta
     \]

4. **Maximize the area of \( \triangle ABC \):**
   - To maximize \( [ABC] \), we need to maximize both \( \sin \theta \) and \( \sin \angle O_1AO_2 \).
   - \( \sin \theta \) is maximized when \( \theta = \frac{\pi}{2} \).
   - \( \sin \angle O_1AO_2 \) is maximized when \( \angle O_1AO_2 = \frac{\pi}{2} \), i.e., \( AO_1 \perp AO_2 \).

5. **Determine the length of \( AD \):**
   - Since \( AO_1 = O_1D = R_1 \) and \( AO_2 = O_2D = R_2 \), \( AO_1DO_2 \) forms a kite, and thus \( AD \perp O_1O_2 \).
   - Let \( X = AD \cap O_1O_2 \).
   - Using the properties of the kite and right triangles, we have:
     \[
     AX = \frac{AO_1 \cdot AO_2}{O_1O_2} = \frac{R_1 R_2}{\sqrt{R_1^2 + R_2^2}}
     \]
   - Therefore:
     \[
     AD = 2 \cdot AX = \frac{2 R_1 R_2}{\sqrt{R_1^2 + R_2^2}}
     \]

The final answer is \(\boxed{\frac{2 R_1 R_2}{\sqrt{R_1^2 + R_2^2}}}\)