= Solution
Substituting $\eta=D_s\beta$ and $\xi=D_d\theta$ into the previous geometric relation verifies
$$
\beta=\theta-\frac{D_{ds}}{D_s}\hat\alpha=\theta-\alpha(\theta).
$$
For a <point mass> <gravitational lens>, the given physical deflection yields
$$
\alpha(\theta)=\frac{4GM}{c^2}\frac{D_{ds}}{D_dD_s}\frac1\theta
=\frac{\theta_E^2}{\theta}.
$$
An aligned source has $\beta=0$, and therefore $\theta^2=\theta_E^2$. <Rotational symmetry> turns these signed intersections into an <Einstein ring>, whose <Einstein radius> is
$$
\boxed{\theta_E=\left(\frac{4GM}{c^2}\frac{D_{ds}}{D_dD_s}\right)^{1/2}.}
$$
For an offset source, multiplying $\beta=\theta-\theta_E^2/\theta$ by $\theta$ gives a <quadratic equation>. Its two image positions are
$$
\boxed{\theta_\pm=\frac{\beta\pm\sqrt{\beta^2+4\theta_E^2}}2.}
$$
For $\beta>0$, $\theta_+>0$ lies on the source side of the lens and $\theta_-<0$ lies on the opposite side.
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