Centre–Chord Perpendicular Bisector

Statement

In a circle, any radius or diameter drawn perpendicular to a chord will bisect that chord. That means the chord is cut into two equal halves at the point of intersection.

Why it’s true (short reason)

• Join the centre of the circle to the endpoints of the chord: you create two congruent right-angled triangles.
• Both triangles share the radius as their hypotenuse, and they share the perpendicular line as a height.
• By symmetry, the two halves of the chord must be equal. This is why the perpendicular radius/diameter bisects the chord.

Recipe (how to use it)

1. Identify the circle, its centre \(O\), and chord \(AB\).
2. Draw a radius or diameter \(OM\) such that \(OM \perp AB\).
3. By the perpendicular bisector property: \(AM = MB\).
4. Use Pythagoras in right triangle \(OAM\) if you need to find chord length from radius and perpendicular distance: \[ OA^2 = OM^2 + AM^2. \]
5. Multiply \(AM\) by 2 to get the full chord length: \(AB = 2AM\).

Spotting it

Use this property when:

• A question involves a perpendicular line from the centre to a chord.
• You are asked for the length of a chord, given the radius and the perpendicular distance from the centre.
• You need to prove that a line bisects a chord or that two segments are equal.

Common pairings

• Pythagoras’ theorem (to calculate chord length or perpendicular distance).
• Circle theorems (equal radii, symmetry of chords).
• Geometry of isosceles triangles (two radii are equal).

Mini examples

1. A radius perpendicular to chord \(AB=12\) cm → each half is \(6\) cm.
2. Circle with radius \(10\) cm, perpendicular from centre to chord is \(6\) cm. By Pythagoras: \(AM = \sqrt{10^2-6^2}=8\). So \(AB = 16\) cm.
3. Circle radius \(13\) cm, perpendicular distance to chord = \(5\) cm. \(AM = \sqrt{13^2-5^2}=12\). So \(AB = 24\) cm.

Pitfalls

• Forgetting that the chord is bisected: A perpendicular radius doesn’t just meet the chord; it splits it into two equal parts.
• Using the whole chord instead of half: In Pythagoras, use \(AM = AB/2\), not the full chord.
• Confusing tangent property: A tangent is perpendicular to a radius at the point of contact — not the same as this theorem.

Exam strategy

• Mark the midpoint \(M\) of the chord where the radius/diameter meets it.
• State clearly: “Radius ⟂ chord ⇒ chord is bisected.”
• Apply Pythagoras: \(OA^2 = OM^2 + AM^2\).
• Multiply the half-chord back up if the full chord length is required.

Extended micro-examples

1. Circle with radius \(17\) cm, perpendicular distance to chord = \(8\) cm. \(AM = \sqrt{17^2-8^2}=\sqrt{225}=15\). So \(AB=30\) cm.
2. Chord length \(40\) cm, perpendicular distance = \(9\) cm. Half chord = \(20\). Radius = \(\sqrt{20^2+9^2}=\sqrt{481}\approx 21.9\) cm.
3. Chord \(AB\) bisected by diameter, \(AM=6\). Then \(AB=12\) cm.

Summary

A radius or diameter drawn perpendicular to a chord always bisects the chord, splitting it into two equal segments. This gives a simple way to find chord lengths or perpendicular distances using Pythagoras’ theorem: \[ AB = 2\sqrt{r^2 - OM^2}. \] Remember: perpendicular → bisected, and the property works for any chord inside the circle.