I’m Up and Down, and Round and Round · Lesson 5 of 8

Distance of Chords from the Centre

“Equal chords stay equally far from the centre because circles appreciate fairness.”

Learning Objectives

• Define the distance of a chord from the centre. • Prove that equal chords are equidistant from the centre. • Prove the converse. • Understand why longer chords lie closer to the centre. • Derive a formula for chord length using radius and perpendicular distance.

The distance from the centre of a circle to a chord means the perpendicular distance. Because the perpendicular from the centre bisects the chord, every chord-distance problem naturally creates a right triangle.

Definition

Distance of a Chord from the Centre

The perpendicular distance from the centre of the circle to the chord.

Equal Chords Are Equidistant from the Centre

Theorem

Chords of a circle having the same length are at the same distance from the centre.

Let AB and CD be equal chords. Drop perpendiculars OM and ON from the centre O. These perpendiculars bisect the chords, so AM=AB/2 and CN=CD/2. Since AB=CD, AM=CN. Also OA=OC because both are radii. The right triangles OMA and ONC are congruent by RHS, giving OM=ON.

Equal chords at equal distances

The Converse

Converse Theorem

Chords of a circle that are equidistant from the centre have equal length.

If OM=ON, then in right triangles OMA and ONC, the hypotenuses OA and OC are equal radii and one leg OM equals ON. RHS congruence gives AM=CN, so doubling both halves gives AB=CD.

Which Unequal Chord Is Closer?

Theorem

Of two unequal chords, the longer chord is closer to the centre.

Suppose AB>CD. Drop perpendiculars OM and ON. Since they bisect the chords, AM>CN. In right triangles, OA and OC are equal radii. By the Baudhāyana–Pythagoras theorem, OA²=OM²+AM² and OC²=ON²+CN². Because the hypotenuse squares are equal but AM²>CN², it follows that OM²<ON², hence OM<ON.

Chord-Length Formula

Let a chord have length L, circle radius r, and perpendicular distance d from the centre. The perpendicular bisects the chord, so half the chord has length L/2. The resulting right triangle gives

Right-triangle relationLaTeX

r^{2} = d^{2} + (\frac{L}{2})^{2}

Chord lengthLaTeX

L = 2 r^{2} - d^{2}

Worked Example: Find Chord Length

Problem
A circle has radius 7 cm and a chord is 6 cm from the centre. Find the chord length.

1.Let half-chord be x.
2.By Pythagoras, 7²=6²+x².
3.49=36+x², so x²=13.
4.x=√13.
5.Full chord length=2√13 cm.

Worked Example: Diameter as the Longest Chord

Problem
Use the chord formula to explain why the diameter is the longest chord.

1.L=2√(r²−d²).
2.The largest value occurs when d is smallest.
3.The smallest possible distance is d=0, when the chord passes through the centre.
4.Then L=2√r²=2r, the diameter.
5.Therefore no chord can be longer than the diameter.

Practice Problems

A chord is 5 cm from the centre of a circle of radius 13 cm. Find its length.
A circle has diameter 26 cm and chord length 24 cm. Find the chord's distance from the centre.
A chord has length 16 cm and is 6 cm from the centre. Find the radius.
Explain why equal chords must have equal perpendicular distances from the centre.
Two chords have lengths 10 cm and 18 cm. Which is closer to the centre? Explain without calculation.
Can we conclude that if one chord is twice as far from the centre as another then the second chord is twice as long? Explain.

Key Takeaways

• Chord distance means perpendicular distance from the centre. • Equal chords are equidistant from the centre, and conversely. • The longer of two unequal chords is closer to the centre. • A perpendicular from the centre bisects a chord and creates a right triangle. • Chord length is 2√(r²−d²). • The diameter is the longest chord.

Coming Next

Next, we move from chords to arcs and connect central angles with angles on the circle.

Previous · Lesson 4

Midpoints and Perpendicular Bisectors of Chords

Next · Lesson 6

Angles Subtended by an Arc