Three non-collinear points determine a unique circle

Depends on: Three perpendicular bisectors meet (circumcenter), Perpendicular bisector ⇔ equidistant from endpoints.

Statement

Let $A$, $B$, $C$ be three non-collinear points in the plane. Then exactly one circle passes through all three; its centre is the circumcenter $O$ of $\triangle ABC$ (the intersection of the three perpendicular bisectors), and its radius is

This is the circumscribed circle of $\triangle ABC$.