Chords Equidistant from Centre

Statement

Chords equidistant from the centre of a circle are equal in length. Conversely, equal chords are equidistant from the centre.

Why it’s true (short reason)

• If you draw a perpendicular from the centre \(O\) to a chord, it bisects the chord (by the centre–chord perpendicular bisector theorem).
• Suppose two chords \(AB\) and \(CD\) are at the same distance from the centre. Then the right-angled triangles formed (e.g. \(OAM\) and \(OCN\)) have:
  • Same hypotenuse (radii of the circle).
  • Same short side (the distance from the centre).
• By Pythagoras, the half-chord lengths are equal, so the full chords are equal.
• Conversely, if chords are equal in length, then their perpendicular distances from the centre must also be equal, by the same reasoning.

Recipe (how to use it)

1. Identify the circle, chords, and distances from centre: If two chords are equally far from the centre, immediately conclude that they have the same length.
2. Use converse: If two chords are equal, you can conclude that their perpendicular distances from the centre are equal.
3. Calculate chord length:
   • Drop a perpendicular from the centre to the chord (bisector).
   • Apply Pythagoras in the right triangle: \(\text{half-chord} = \sqrt{r^2 - d^2}\), where \(r\) = radius, \(d\) = distance from centre.
   • Chord length \(= 2 \times \text{half-chord}\).

Spotting it

This theorem is useful when:

• You are told that two chords are at the same distance from the centre.
• You need to prove two chords are equal in length.
• Equal chord lengths are given and you are asked to find or prove equal distances from the centre.

Common pairings

• Pythagoras’ theorem to calculate chord length or distance from centre.
• Perpendicular bisector theorem (radius ⟂ chord → bisects chord).
• Circle symmetry arguments in geometry proofs.

Mini examples

1. Chords \(AB\) and \(CD\) are each 5 cm from the centre. If \(AB = 14\) cm, then \(CD = 14\) cm.
2. A circle of radius 10 cm has a chord 6 cm from the centre. Half-chord = \(\sqrt{10^2 - 6^2} = \sqrt{64} = 8\). So full chord = 16 cm. Another chord 6 cm away also has length 16 cm.
3. Two chords are each 20 cm long. Therefore, both are equidistant from the centre.

Pitfalls

• Forgetting to halve the chord: Pythagoras applies to half the chord, not the whole.
• Mixing up converse and direct statement: Remember both directions: equidistant chords are equal, and equal chords are equidistant.
• Unit mistakes: Distances are linear (cm, m), not squared units.

Exam strategy

• Underline keywords: “equidistant” or “equal chords.”
• If calculation needed, sketch triangle with radius, half-chord, and distance from centre.
• Use formula: \(\text{Chord length} = 2\sqrt{r^2 - d^2}\).
• Check your result makes sense: chord length should be less than the diameter.

Extended micro-examples

1. Radius 13 cm, chord 5 cm from centre: Half-chord = \(\sqrt{13^2 - 5^2} = \sqrt{144} = 12\). So chord = 24 cm. Another chord at 5 cm distance also = 24 cm.
2. Chord 30 cm long, distance from centre 4 cm: Half-chord = 15. Radius = \(\sqrt{15^2 + 4^2} = \sqrt{225+16} = \sqrt{241} \approx 15.5\) cm.
3. Two chords equal (18 cm each): They are equidistant from the centre, so both are at the same perpendicular distance.

Summary

Key idea: In a circle, chords at equal distances from the centre are equal in length, and equal chords are equidistant from the centre. Use Pythagoras with the radius and perpendicular distance to calculate chord lengths or distances. This theorem underpins many circle proofs and problem-solving strategies in GCSE geometry.