Orthogonality of reciprocal-coordinate circles ($\mathrm{Re}(1/z)=a,\mathrm{Im}(1/z)=b$)

For nonzero real $a,b$, the two conditions describe circles $(x-1/(2a){)}^2+y^2=1/(4a^2)$ and $x^2+(y+1/(2b){)}^2=1/(4b^2)$, respectively, with the origin removed because $1/z$ is undefined there. Their unique actual intersection is $z=(a-ib)/(a^2+b^2)$. The map $z\mapsto 1/z$ is conformal away from zero and maps these circles to perpendicular coordinate lines, so they intersect orthogonally. The excluded origin is an intersection of their closures, not of the original loci. For $a=b=1$, both lie inside the unit circle and touch it at $1$ and $-i$.