 One-point value region of a holomorphic disc map

For holomorphic functions $f:\mathbb{D}\to \mathbb{D}$ with $f(z_0)=w_0$, put $r=|(z-z_0)/(1-{\overline{z}}_{0}z)|$. Their possible values at $z$ form the closed disc with centre $(1-r^2)w_0/(1-r^2|w_0{|}^2)$ and radius $r(1-|w_0{|}^2)/(1-r^2|w_0{|}^2)$. Indeed ${\psi }_{w_0}\circ f={\psi }_{z_0}h$ with holomorphic $|h|\le 1$; all constant choices of $h$ in the closed unit disc are admissible. Completing the square in the displayed pseudohyperbolic inequality gives the centre and radius.