Brownian exit law from a quadrant ($q_a(r)=\frac{4a^{2}r}{\pi (r^4+4a^4)}$)

Planar Brownian motion started at $a+ia$, $a>0$, exits the positive quadrant through either positive coordinate axis with probability one-half. The density along each axis is $q_a(r)$ for $r>0$, measured with respect to distance along that axis. Squaring maps the quadrant to the upper half-plane and sends the starting point to $2i{a}^2$. The boundary square therefore has the centered Cauchy distribution of scale $2a^2$; pulling it back by $r\mapsto r^2$ gives the displayed density. The radius has distribution function $2\arctan (r^2/(2a^2))/\pi$, independently of which axis is chosen. There is no atom at the origin.