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Geometry & measures · Circles, arcs & sectors

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Tangents, chords & the alternate segment

Why a tangent meets a radius at 90°, why two tangents from a point are equal, why the perpendicular from the centre bisects a chord, and the alternate segment theorem.

Geometry & measures · Circles, arcs & sectors

Tangents, chords & the alternate segment

Why a tangent meets a radius at 90°, why two tangents from a point are equal, why the perpendicular from the centre bisects a chord, and the alternate segment theorem.

Why it works

A tangent is perpendicular to the radius at the point of contact. A tangent touches the circle exactly once, so every other point on it lies outside the circle — further from the centre. The point of contact is therefore the closest point of the tangent to the centre, and the shortest distance from a point to a line is always the perpendicular. So the radius meets the tangent at 90°90°.

Two tangents from the same external point are equal. From a point PP outside the circle, draw both tangents, touching at AA and BB, and join OPOP. Triangles OAPOAP and OBPOBP each have a right angle (radius–tangent), share the hypotenuse OPOP, and have equal sides OA=OBOA = OB (radii) — so they are congruent by RHS. Hence PA=PBPA = PB, and OPOP bisects both the angle APBAPB and the angle AOBAOB. That kite shape, symmetric about OPOP, is worth recognising on sight.

The perpendicular from the centre bisects a chord. Drop a perpendicular from OO to a chord ABAB, meeting it at MM. Triangles OAMOAM and OBMOBM have a right angle, the common side OMOM, and OA=OBOA = OB (radii) — congruent by RHS again, so AM=MBAM = MB. This is the tool for chord-length questions: the radius, half the chord and the distance from the centre make a right-angled triangle, so Pythagoras finishes the job.

The alternate segment theorem. The angle between a tangent and a chord equals the angle in the alternate segment — the angle subtended by that chord from the other side of it. It follows from the earlier theorems: the tangent–chord angle and the angle at the centre both relate to the same arc, and the radius–tangent right angle converts one into the other. In practice, spot it by looking for a tangent, a chord leaving its point of contact, and a triangle inscribed on the far side of that chord.

Reasons, in full, again:
  • "the angle between a tangent and a radius is 90°"
  • "tangents from an external point are equal"
  • "the perpendicular from the centre to a chord bisects the chord"
  • "the angle between a tangent and a chord equals the angle in the alternate
segment"