Chord Properties

A chord is any line segment whose two endpoints both lie on a circle. The simplest example is a diameter, which passes through the centre — but most chords sit to one side. One of the most elegant results in Euclidean geometry connects just two measurements: the length of a chord and the perpendicular distance from the centre of the circle to that chord. Understanding this relationship unlocks a family of powerful theorems used in constructions, engineering, and proof-writing. The key insight is that the perpendicular from the centre of a circle to a chord always bisects that chord — cutting it into two equal halves. This is not a coincidence; it follows directly from the symmetry of the circle and can be proved rigorously using congruent triangles. Conversely, if you know a line from the centre bisects a chord, that line must be perpendicular to the chord. These two directions together form a biconditional theorem that is central to circle geometry. Chord properties also reveal how distance from the centre governs length: chords that are closer to the centre are longer, while chords farther away are shorter. Equal chords are equidistant from the centre, and chords at equal distances from the centre are equal in length. These facts let us compare chords without ever measuring them directly, making them indispensable tools in deductive geometric reasoning.

Key takeaways
  • Set up the diagram
  • Identify the right triangles
  • Conclude bisection
  • Apply the Pythagorean theorem
  • Equal chords theorem
  • Longer chords lie closer to the centre

Set up the diagram

Let circle with centre O have radius r. Draw chord AB. Drop a perpendicular from O to AB, meeting AB at point M. By definition, OM ⊥ AB, so angle OMA = 90°.

Identify the right triangles

Since OA and OB are both radii, OA = OB = r. The perpendicular OM is common to triangles OMA and OMB. Both triangles share the right angle at M. Therefore triangles OMA and OMB are congruent by the Hypotenuse-Leg (HL) theorem: hypotenuse OA = OB and leg OM is shared.

Conclude bisection

From the congruence OMA ≅ OMB, corresponding sides give MA = MB. This proves the perpendicular from the centre bisects the chord. Denoting the half-chord length as AM = l/2 (where l = AB), and the distance OM = d, we have a right triangle with legs d and l/2, and hypotenuse r.

Apply the Pythagorean theorem

In right triangle OMA: OA² = OM² + AM², which gives r² = d² + (l/2)². Rearranging: l² = 4(r² − d²), so l = 2√(r² − d²). This is the fundamental chord–distance formula.

Equal chords theorem

If two chords AB and CD have equal lengths (AB = CD), then their distances from the centre must be equal: d₁ = d₂. Proof: from l = 2√(r² − d²), the same value of l with the same r gives the same value of d. The converse holds identically — equal distances produce equal chord lengths.

Longer chords lie closer to the centre

Since l = 2√(r² − d²), increasing d decreases the value under the square root, so l decreases. The diameter (d = 0) is the longest possible chord: l_max = 2r. A chord with d = r would have length 0 — it degenerates to a single point (a tangent contact). This confirms the inverse relationship between distance and chord length.

The Key Equation

l=2r2−d2l = 2\sqrt{r^{2} - d^{2}}
  • l — Chord length (length units): The full length of the chord from one endpoint on the circle to the other.
  • r — Radius (length units): The radius of the circle; the distance from the centre to any point on the circumference.
  • d — Perpendicular distance (length units): The shortest distance from the centre O to the chord, measured along the perpendicular from O to the chord. Always satisfies 0 ≤ d ≤ r.

Worked Example

Problem. A circle has centre O and radius r = 13 cm. A chord AB is drawn such that the perpendicular distance from O to AB is d = 5 cm. Find the length of chord AB. Then find the distance from O to a second chord CD that has length 10 cm.

AB=24 cm,dCD=12 cmAB = 24\text{ cm},\quad d_{CD} = 12\text{ cm}

Answer. Chord AB has length 24 cm; the perpendicular distance from O to chord CD is 12 cm.

Figure
Diagram
OABMdrrl/2l/2Chord–Distance TheoremOM ⊥ AB ⟹ MA = MB (M is midpoint of AB)r² = d² + (l/2)² ⟹ l = 2√(r² − d²)Chord ABPerpendicular distance dRadii OA = OB = r
Origins · 1 minEuclid of Alexandria and how this was discovered
EO
Euclid of Alexandria
-325–-265 · Greek

Euclid systematised Greek geometry in his thirteen-book treatise Elements (c. 300 BCE). Book III is devoted entirely to the geometry of circles and contains the formal proofs of chord and perpendicular bisector theorems that students still study today. His deductive method — beginning from definitions, postulates, and common notions — remains the model for all rigorous mathematics. When King Ptolemy I asked Euclid if there was a shortcut to learning geometry, Euclid reportedly replied: 'There is no royal road to geometry.' The phrase has been used ever since to mean that hard work cannot be bypassed.

  • Tangent Properties — A tangent is the limiting case of a chord as distance d approaches r, making chord length approach zero. The radius-to-tangent perpendicularity theorem uses the same right-triangle reasoning.
  • Arc–Chord Relationship — Equal chords subtend equal arcs (and vice versa), directly extending the equal-chords theorem to arc measure and central angles.
  • Pythagorean Theorem — The chord length formula l = 2√(r² − d²) is derived by applying the Pythagorean theorem to the right triangle formed by the radius, perpendicular distance, and half-chord.
  • Triangle Congruence — The proof that the perpendicular from the centre bisects the chord relies on the Hypotenuse-Leg congruence condition for right triangles.