Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 1 / 1B / a / Solution

Translate and scale the circle to $|z|=1$. For distinct points $u,v,w$ on it, the angle at $w$ is right exactly whenis purely imaginary. Since $\bar{u}=u^{-1}$ and similarly for $v,w$, direct simplification gives $\bar{R}=(v/u)R$. Thus $\bar{R}=-R$ exactly when $v=-u$, which says that the opposite side joins antipodal points and is a diameter. The converse is the same calculation in reverse.