Answer: If three points of a circle were collinear, the centre would lie on two distinct parallel perpendicular bisectors, which is impossible.
Suppose, if possible, that three distinct points A, B, C on a circle with centre O are collinear, with A–B–C.
- OA = OB (radii), so O lies on the perpendicular bisector of seg AB.
- OB = OC (radii), so O lies on the perpendicular bisector of seg BC.
Both perpendicular bisectors are perpendicular to the same line ABC, so they are parallel. They pass through the midpoints of AB and BC, which are different points, so they are two distinct parallel lines and have no common point.
This contradicts that O lies on both. Hence our assumption is wrong: any three points on a circle cannot be collinear.