260. The Unique Circle · Three points, one circle, no more, no less

Try to break it

Drag A, B, or C anywhere — as long as they don't all line up. The three perpendicular bisectors always meet at a single point O, and O is equidistant from the three vertices, so a single circle passes through them. Drag the three points onto one straight line and the bisectors become parallel — the circle vanishes (radius → ∞).

How you build it

Construct a circle through three points.

• Mark point A — the first of three non-collinear points.
• Mark point B.
• Mark point C — make sure it is NOT on line AB.
• Draw segment AB.
• Draw segment BC.
• Construct the perpendicular bisector of AB — every point on it is equidistant from A and B.
• Construct the perpendicular bisector of BC — every point on it is equidistant from B and C.
• Mark O at the intersection of the two perpendicular bisectors — this is the unique circumcentre, equidistant from A, B and C.
• With centre O and radius OA, draw the unique circle through A, B and C.

The proof, step by step

Prove that the perpendicular bisectors of the sides of a triangle meet at one point equidistant from the vertices.

• O lies on the perpendicular bisector of AB. Therefore, OA = OB.
• O lies on the perpendicular bisector of BC. Therefore, OB = OC.
• From (1) and (2), OA = OB = OC. So O is equidistant from A, B, and C.
• Two non-parallel lines intersect at exactly one point. Thus, O is unique, and only one circle passes through A, B, and C.

Worked example

In triangle ABC, the perpendicular bisectors of the sides meet at point O. If the distance OA is 10 cm, what is the radius of the circumcircle of triangle ABC?

By Theorem 26, the perpendicular bisectors of a triangle's sides meet at the circumcenter O, which is equidistant from all vertices. Since OA = 10 cm, the radius of the circumcircle is exactly 10 cm.

• 5 cm
• 10 cm — correct
• 15 cm
• 20 cm