Observe that when $AD$ is a certain fixed value, $\angle B, \angle C$ (or their supplementary angles, hereafter the same) are fixed angles, and therefore $\angle A$ is a fixed angle. Since
$$\begin{aligned} |\triangle ABC| &= \frac{1}{2} AB \cdot AC \sin \angle A \\ &= 2R_1 R_2 \sin \angle ADB \sin \angle ADC \sin \angle A \\ &= 2R_1 R_2 \sin^2 \angle ADB \sin \angle A. \end{aligned}$$
the area attains its maximum value when $\angle ADB = 90^\circ$. When $AD$ is allowed to vary, this value attains its maximum when $\angle A = 90^\circ$. And when the two circles are orthogonal, that is, when $\angle O_1AO_2 = 90^\circ$, we have $\angle A = 90^\circ$. Let $AD = l$ at this time, then
$$\begin{aligned} l^2 &= \sqrt{4R_1^2 - l^2} \sqrt{4R_2^2 - l^2} \\ \Rightarrow \quad l^4 &= (4R_1^2 - l^2)(4R_2^2 - l^2) = 16R_1^2 R_2^2 - 4(R_1^2 + R_2^2)l^2 + l^4, \end{aligned}$$
therefore
$$l = \frac{2}{\sqrt{\frac{1}{R_1^2} + \frac{1}{R_2^2}}}$$