Chord of Circle Length Theorem and Formula

A circle is a round shaped figure that is a combination of many points in a plane. The points are at a constant distance (radius) from the fixed center point of a circle. A chord of the circle is the straight line segment which has both its endpoint on the circle. The chord of a circle that passes through the center of the circle divides the circle into two halves, and is the longest chord of a circle, also called the diameter of a circle.

Key takeaways: Chord of a circle, Formula, Length of chord of a circle, Theorems, Circle

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Length of Chord of a Circle Formula

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Length of chord(trigonometry) = 2 × (r²-d²)

Length of a chord using perpendicular distance = 2 × r × sin (c/2)

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Theorem of Chord of a Circle with Proof

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Theorem 1: Both the Equal chords of a circle that subtends to an Equal angle at the center of a circle

Proof: A circle has two equal chords AC and BD of equal length. The two chords form a two triangle with a circle center O and we want to prove that ∠ AOC = ∠ BOD

Solution: The two chords AC and BD form a two circle with the centre as ∆AOB & ∆COD

In triangles AOC and BOD,

The radius OA=OB & OC=OD

Given, AC=BD

Therefore by the SSS rule, ∆AOC≅∆BOD

This proves ∠ AOC = ∠ BOD

(SSS Rule- If all the sides of a triangle are equal to all the sides of another triangle, then the triangles are said to be congruent (all corresponding angle pairs are equal))

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Theorem 2: The Angles subtended by chords at the center of the circle are equal, then the length of the two chords are equal.

Proof: A Circle has two chords AC and BD. The angles formed by these two chords at the center of a circle is the same (This theorem is the converse of Theorem 1).

Solution: Reference with Theorem 1 diagram If ∠ AOC = ∠ BOD

by SSS rule, ∆AOC≅∆BOD

so, OA=OB & OC=OD

This proves AC=BD

Theorem 3: The perpendicular line from the center of the chord bisects the chord.

Proof: AB is a chord of the circle with a center at O and the perpendicular line passes through the circle center O and cuts the chord into two halves.

it results in two right-angled triangles ∆AOM & ∆BOM, so ∠ AOM = ∠ BOM=90° 

OA=OB(radius of the circle)

By SSS rule, ∆AOB≅∆COD

This proves that AM=BM, the perpendicular line bisects the chord.

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Theorem 4: A Line passes through the center of the circle bisects the chord is perpendicular to the chord (This theorem is the converse of theorem 3)

Proof: AB is the chord of a circle that is bisected by a line from center O

OA = OB (Radius of the circle are equal)

AM = BM(Line bisects the chord)

By SSS Rule, ∆AOB ≅ ∆COD

This proves that ∠ AOM = ∠ BOM=90°, so the line bisects the chord is perpendicular to the chord.

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Theorem 5: Both the Equal chord of a circle are at equal distance from the center of a circle

Proof: AB and CD are two chords of a circle with center O and we have to prove that AB and CD are at equal distances from center O

The distance of AB from the center is marked as X and The distance of CD from the center is marked as Y

X bisects the AB as it is perpendicular to AB and as the same Y bisects the CD as it is perpendicular to CD

so AX=BX = CY=DY

We take, ∆AOX & ∆COY

AX=CY (perpendicular bisectors)

OA=OC(Radius of a circle are same)

By SSS rule ∆AOX≅∆COY and by RHS criterion, ∠ AOX = ∠ COY=90°,

This proves that OX=OY, so two chords are at equal distance from the center O

Theorem 6: Two Chords of a circle which are at the same distance from the Center O are equal in length (Converse of Theorem 5)

Proof: AB and CD are two chords of a circle with center O, Both chords are at an equal distance from the center O, we have to prove that AB=CD

Regarding the details in the above theorem,

OX=OY(Distance from the center are equal)

OA=OC(radius of a circle are same)

By SSS rule, ∆AOX≅∆COY and by RHS criterion, ∠ AOX = ∠ COY=90°,

This proves that AX=CY & BX=DX

So, AB=CD the given two chords are equal

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Circle Formulas

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• Area of a circle: A = πr²
• Diameter of a circle: D = 2r
• Circumference of a circle: C = 2πr

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Things to Remember

• A circle is a combination of points in a place that is at equidistant from the center O.
• A chord of a circle is a line that joins the two points on the boundary of the circle and it is called a secant if the chord is extended external to the circle. 
• The line that bisects the chord from the center of the circle is perpendicular to the chord and that line is referred to as the perpendicular bisector of the chord.
• Chords which are having the same length are at equidistant from the center of the circle and the chords which are at equidistant from the center are of equal length.

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