Chord of a Circle: Formula, Theorems, Examples

Chord of a Circle: Formula, Theorems, Examples

TL;DR

A chord of a circle is a straight line segment joining any two points on the circle's boundary, and the longest possible chord is the diameter. This article covers the definition, the two formulas for chord length (from the perpendicular distance and from the central angle), the main chord theorems with proof, six worked examples, and the common mistakes.

Every Straight Cut Across a Circle Has the Same Hidden Rule

Slice a round cake anywhere with a straight knife, not necessarily through the middle, and the cut you leave on the top is a chord. Make the cut nearer the edge and it is short; slide it toward the centre and it grows. Push it right through the middle and you get the longest cut the cake allows. Behind that simple picture sits a single right-triangle relationship that fixes the length of every chord from just two numbers.

Once you can see the radius, half the chord, and that perpendicular forming a right triangle, every chord-length problem becomes the Pythagorean theorem in disguise.

What Is a Chord of a Circle?

A chord of a circle is a straight line segment whose two endpoints both lie on the circle's boundary. Join any two points on the circumference with a straight line, and that segment is a chord.

The diameter is a special chord: it is the chord that passes through the centre, and it is the longest chord any circle has. Every other chord is shorter, because no straight line across a circle can be longer than the one that goes through the middle. A chord also splits the circle into two pieces, called segments, a larger major segment and a smaller minor segment.

It helps to place the chord alongside the other circle parts. The radius runs from the centre to the boundary; a chord runs from boundary to boundary; the diameter is the chord that happens to pass through the centre. A short menu of these terms sits in parts of a circle, but the one feature that defines a chord is simple: both ends touch the circle, and it need not go through the centre.

The Chord Length Formula — Two Ways to Find It

The length of a chord can be found two ways, depending on what the problem gives you. Define the variables first: L is the chord length, r the radius of the circle, d the perpendicular distance from the centre to the chord, and θ the angle the chord subtends at the centre.

From the perpendicular distance.

Drop a perpendicular from the centre to the chord. It hits the chord at its midpoint (that is a theorem we prove below), so it splits the chord into two equal halves. The radius, one half-chord, and the perpendicular form a right triangle, with the radius as the hypotenuse. By the Pythagorean theorem, the half-chord is ( \sqrt{r^2 - d^2} ), and the full chord is twice that:

[ L = 2\sqrt{r^2 - d^2} ]

From the central angle.

If instead you know the angle ( \theta ) the chord subtends at the centre, the chord is the base of an isosceles triangle with two radius sides. Splitting that triangle down its line of symmetry gives a right triangle whose opposite side is half the chord, so the half-chord is ( r \sin \left( \frac{\theta}{2} \right) ), and the full chord is:

[ L = 2r\sin \left( \frac{\theta}{2} \right) ]

Both formulas describe the same chord; you pick whichever matches the information you are handed. If the angle is given, use the second; if a perpendicular distance is given, use the first.

The Chord Theorems — and Why They Hold

A handful of theorems about chords come up again and again, and most school problems are an application of one of them. Here are the three that matter most.

Theorem 1: The perpendicular from the centre bisects the chord

If you drop a perpendicular from the centre of a circle to a chord, it cuts the chord exactly in half. The proof is short: join the centre to both ends of the chord, giving two radii of equal length; the perpendicular is shared by the two triangles formed, and each has a right angle where it meets the chord. The two triangles are congruent (right angle, hypotenuse, side), so the two half-chords are equal. This is the theorem the length formula quietly relies on.

Theorem 2: Equal chords are equidistant from the centre, and chords equidistant from the centre are equal

Two chords of the same length sit the same perpendicular distance from the centre, and the rule works in reverse too. This follows straight from the length formula: ( L = 2\sqrt{r^2 - d^2} ) depends only on ( r ) and ( d ), so for a fixed radius, equal ( L ) forces equal ( d ).

Theorem 3: The larger of two unequal chords is closer to the centre

Look at the formula again: as the perpendicular distance ( d ) shrinks toward zero, the chord length ( L ) grows. When ( d = 0 ) the chord passes through the centre and becomes the diameter, the longest chord. So the closer a chord sits to the centre, the longer it is.

These three are not separate facts to memorise; they are all the single relationship ( L = 2\sqrt{r^2 - d^2} ) read in different directions. Equal chords, equidistant chords, the longest chord, each is what the formula says when you hold one quantity fixed and vary another.

Examples of the Chord of a Circle

With the formulas and theorems in place, here is the chord doing real work. The problems move from a one-step length calculation up to a proof-style application of the bisection theorem.

Example 1: A chord lies 4 cm from the centre of a circle whose radius is 5 cm. Find the length of the chord

[ L = 2\sqrt{r^2 - d^2} = 2\sqrt{25 - 16} = 2\sqrt{9} = 6 \text{ cm} ]

Example 2: A chord subtends an angle of 60° at the centre of a circle of radius 10 cm. Find the chord length

[ L = 2r\sin \left( \frac{\theta}{2} \right) = 2(10)\sin 30° = 10 \text{ cm} ]

Example 3: The perpendicular distance from the centre to a chord is 8 cm, and the chord is 12 cm long. Find the radius of the circle

[ r^2 = \left( \frac{L}{2} \right)^2 + d^2 = 6^2 + 8^2 = 100 \implies r = 10 \text{ cm} ]

Example 4: Two parallel chords of a circle of radius 13 cm have lengths 10 cm and 24 cm. Find the distance between them if they lie on the same side of the centre

[ d_1 = \sqrt{13^2 - 5^2} = 12 \text{ cm}, \quad d_2 = \sqrt{13^2 - 12^2} = 5 \text{ cm} ]
The gap between them is: 12-5=7 cm.

Example 5: A chord of a circle is equal in length to the radius. Find the angle it subtends at the centre

The triangle is equilateral, so the central angle is 60°.

Example 6: Prove that the perpendicular from the centre of a circle to a chord bisects the chord

Join OA and OB; both are radii, so ( OA = OB ). The perpendicular from O meets AB at M, so ( \angle OMA = \angle OMB = 90° ). Triangles OMA and OMB share the side OM and have equal hypotenuses, so they are congruent. Thus, M is the midpoint and the perpendicular bisects the chord.

Where Chords Show Up

Chords matter because they turn a curved boundary into a straight, measurable line. This is exactly what engineers and designers need to work with a circle.

Where Students Trip Up on Chords

Mistake 1: Using the full central angle instead of half.

The correct way: Splitting the isosceles triangle of two radii uses half the central angle.

Mistake 2: Forgetting that the perpendicular bisects the chord.

Correct way: The perpendicular from the centre cuts the chord in half.

Mistake 3: Confusing the perpendicular distance with the radius.

Correct way: Keep the right triangle clear: the radius is the hypotenuse, and the perpendicular distance is one leg.

Key Takeaways