Chord Of A Circle, Its Length and Theorems

Chord of a Circle Definition

The chord of a circle can be defined as the line segment joining any two points on the circumference of the circle. It should be noted that the diameter is the longest chord of a circle which passes through the center of the circle. The figure below depicts a circle and its chord.

In the given circle with ‘O’ as the center, AB represents the diameter of the circle (longest chord), ‘OE’ denotes the radius of the circle and CD represents a chord of the circle.

Let us consider the chord CD of the circle and two points P and Q anywhere on the circumference of the circle except the chord as shown in the figure below. If the endpoints of the chord CD are joined to the point P, then the angle ∠CPD is known as the angle subtended by the chord CD at point P. The angle ∠CQD is the angle subtended by chord CD at Q. The angle ∠COD is the angle subtended by chord CD at the center O.

Chord Length Formula

There are two basic formulas to find the length of the chord of a circle which are:

Where,

• r is the radius of the circle
• c is the angle subtended at the center by the chord
• d is the perpendicular distance from the chord to the circle center

Example Question Using Chord Length Formula

Question: Find the length of the chord of a circle where the radius is 7 cm and perpendicular distance from the chord to the center is 4 cm.

Solution:

Given radius, r = 7 cm

and distance, d = 4 cm

Chord length = 2√(r²−d²)

⇒ Chord length = 2√(7²−4²)

⇒ Chord length = 2√(49−16)

⇒ Chord length = 2√33

⇒ Chord length = 2×5.744

Or , chord length = 11.48 cm

Video Related to Chords

Chord of a Circle Theorems

If we try to establish a relationship between different chords and the angle subtended by them in the center of the circle, we see that the longer chord subtends a greater angle at the center. Similarly, two chords of equal length subtend equal angle at the center. Let us try to prove this statement.

Theorem 1: Equal Chords Equal Angles Theorem

Statement: Chords which are equal in length subtend equal angles at the center of the circle.

Proof:

From fig. 3, In ∆AOB and ∆POQ

Note: CPCT stands for congruent parts of congruent triangles.

The converse of theorem 1 also holds true, which states that if two angles subtended by two chords at the center are equal then the chords are of equal length. From fig. 3, if ∠AOB =∠POQ, then AB=PQ. Let us try to prove this statement.

Theorem 2: Equal Angles Equal Chords Theorem (Converse of Theorem 1)

Statement: If the angles subtended by the chords of a circle are equal in measure, then the length of the chords is equal.

Proof:

From fig. 4, In ∆AOB and ∆POQ

Theorem 3: Equal Chords Equidistant from Center Theorem

Statement: Equal chords of a circle are equidistant from the center of the circle.

Proof:

Given: Chords AB and CD are equal in length.

Construction: Join A and C with centre O and drop perpendiculars from O to the chords AB and CD.

Solved Examples

Example 1:

A chord of a circle is equal to its radius. Find the angle subtended by this chord at a point in the major segment.

Solution:

Let O be the centre, and AB be the chord of the circle. 

So, OA and OB be the radii.

Given that chord of a circle is equal to the radius.

AB = OA = OB

Thus, ΔOAB is an equilateral triangle. 

That means ∠AOB = ∠OBA = ∠OAB = 60° 

Also, we know that the angle subtended by an arc at the centre of the circle is twice the angle subtended by it at any other point in the remaining part of the circle. 

So, ∠AOB = 2∠ACB 

⇒ ∠ACB = 1/2 (∠AOB) 

⇒ ∠ACB = 1/2 (60°) 

= 30°

Hence, the angle subtended by the given chord at a point in the major segment is 30°.

Example 2:

Two chords AB and AC of a circle subtend angles equal to 90º and 150º, respectively at the centre. Find ∠BAC, if AB and AC lie on the opposite sides of the centre.

Solution:

Given,

Two chords AB and AC of a circle subtend angles equal to 90º and 150º.

So, ∠AOC = 90º and∠AOB = 150º

In ΔAOB,

OA = OB (radius of the circle)

As we know, angles opposite to equal sides are equal.

So, ∠OBA = ∠OAB

According to the angle sum property of triangle theorem, the sum of all angles of a triangle = 180°

In ΔAOB, 

∠OAB + ∠AOB +∠OBA = 180°

∠OAB + 90° + ∠OAB = 180°

2∠OAB = 180° – 90°

2∠OAB = 90°

⇒ ∠OAB = 45°

Now, in ΔAOC,

OA = OC (radius of the circle)

As mentioned above, angles opposite to equal sides are equal.

∴ ∠OCA = ∠OAC

Using the angle sum property in ΔAOB, we have;

∠OAC + ∠AOC +∠OCA = 180°

∠OAC + 150° + ∠OAC = 180°

2∠OAC = 180° – 150°

2∠OAC = 30°

⇒ ∠OAC = 15°

Now, ∠BAC = ∠OAB + ∠OAC

= 45° + 15°

= 60°

Therefore, ∠BAC = 60°

Practice Problems

1. Two equal chords AB and CD of a circle, when produced, intersect at a point P. Prove that PB = PD.
2. Two circles of radii 5 cm and 3 cm intersect at two points, and the distance between their centres is 4 cm. Find the length of the common chord.
3. The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at a distance of 4 cm from the centre, what is the distance of the other chord from the centre?

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