A Circle Through Three Points

≈ 32 min

A Circle Through Three Points

Given any three points that are not in a straight line, exactly one circle passes through all three. Finding its centre is a direct application of the perpendicular bisector.

The key idea: the centre must be equidistant from all three points. The set of points equidistant from A and B is the perpendicular bisector of AB. The set equidistant from B and C is the perpendicular bisector of BC. Where those two bisectors cross is the only point equidistant from all three — the centre. Set the compasses from there to any of the points and draw.

Worked example. This is how a broken circular part is reconstructed. Mark any three points on the surviving arc, bisect two of the chords between them, and their crossing gives the original centre — so the original radius can be recovered from a fragment.

Core checkpoint: Two bisectors are enough. The third would cross at the same point and only serves as a check.

Why does the crossing of the perpendicular bisectors of two chords give the centre of the circle?

Every point on a chord's perpendicular bisector is equidistant from that chord's two ends, so the crossing is equidistant from all three points.

Because bisectors always cross at the middle of the page.

Because the chords are equal in length.

Because the bisectors are parallel to the radius.

What is the minimum number of chord perpendicular bisectors you must construct to locate the centre?

What happens if the three given points lie in a straight line?

No circle can pass through them — the bisectors are parallel and never cross.

Infinitely many circles pass through them.

The circle has its centre at the middle point.

The construction still works but gives a very small circle.

A fitter has a broken circular flange with only part of the rim surviving. How can the original radius be recovered?

Mark any three points on the surviving arc, bisect two of the chords between them, and the crossing gives the original centre.

Measure the longest surviving piece and halve it.

The radius cannot be recovered from a fragment.

Measure the thickness of the rim.