 Solution

Substituting $\eta =D_s\beta$ and $\xi =D_d\theta$ into the previous geometric relation verifiesFor a point mass gravitational lens, the given physical deflection yieldsAn 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 isFor an offset source, multiplying $\beta =\theta -{\theta }_E^2/\theta$ by $\theta$ gives a quadratic equation. Its two image positions areFor $\beta >0$, ${\theta }_{+}>0$ lies on the source side of the lens and ${\theta }_{-}<0$ lies on the opposite side.