Circles

331. Equal Chords, Equal Distances · why symmetry keeps chords balanced from the centre

OM and ON are always equal in length, no matter where the chords sit.

OMNBDOM = 181.6OM = 181.6ON = 181.6ON = 181.6AC
Equal chords of a circle are equidistant from the centre, and conversely chords equidistant from the centre are equal. So when two chords have the same length, the perpendiculars OM and ON from the centre are equal.

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Selina ICSE: Circles

What this lesson covers

Try to break it

Drag A and C around. Both chords stay equal in length, and the perpendicular distances OM (to AB) and ON (to CD) always come out the same. Equal chords ⇔ equidistant from the centre. Try to find a position where OM ≠ ON; impossible.

How you build it

Draw two equal chords and show their perpendicular distances from O are equal.

  • Mark O near the middle — the centre of the circle.
  • Click O as the centre, then click outward to set the radius.
  • Mark A on the circle.
  • Mark B elsewhere on the circle — AB is the first chord.
  • Join A to B — the first chord.
  • Mark C on the circle, away from A and B — the start of the second chord.
  • Click A as the centre, then click B — the compass opens to exactly the chord length AB and locks it.
  • With the compass still set to AB, click C — swing an arc that crosses the circle. The crossing is exactly AB away from C.
  • Mark D where the arc crosses the circle. Now CD equals AB — a second, equal chord.
  • Join C to D — the second chord, equal in length to AB.
  • From O, drop the perpendicular to chord AB. Its foot is M and OM is the distance of AB from the centre.
  • From O, drop the perpendicular to chord CD. Its foot is N. Since CD = AB, the distance ON equals OM — equal chords are equidistant from the centre.

The proof, step by step

Prove that equal chords of a circle are equidistant from the centre.

  • In ΔOMA and ΔONC, OA = OC (Radii of the same circle).
  • ∠OMA = ∠ONC = 90° (OM ⊥ AB and ON ⊥ CD by construction).
  • AB = CD (Given), so AM = CN (Perpendicular from centre bisects the chord).
  • ΔOMA ≅ ΔONC (RHS congruence criterion).
  • Therefore, OM = ON (c.p.c.t.). Equal chords are equidistant from the centre.

Worked example

In a circle of radius 10 cm, two equal chords of length 12 cm are drawn. Find the distance of each chord from the centre.

Half chord length = 12/2 = 6 cm. Radius = 10 cm. Distance from centre = √(10² - 6²) = √(100 - 36) = √64 = 8 cm.

  • 4 cm
  • 6 cm
  • 8 cm — correct
  • 10 cm
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