MathematicsCore20 min read

Chords and Arcs of a Circle

What the centre knows about every chord drawn in the circle

This topic appears in:

10th Class Mathematics · ch.8O Level Mathematics (4024) · ch.4
01

The vocabulary, settled first

Circle questions are lost more often to confused vocabulary than to confused geometry, so it is worth being exact. A chord is a straight line joining two points on the circle. A diameter is a chord through the centre, and it is the longest chord there can be. An arc is a piece of the circumference; two points cut the circle into a minor arc and a major arc.

A chord also divides the interior into two segments, minor and major. Two radii cut off a sector, which is the pie-slice shape bounded by two radii and an arc — a segment is bounded by a chord, a sector by two radii, and confusing them costs marks.

TermBounded byNote
Chordtwo points on the circlethe diameter is the longest one
Arcpart of the circumferenceminor if less than half, major if more
Segmenta chord and an arcthe part of the interior cut off by a chord
Sectortwo radii and an arcthe pie slice; a semicircle is a sector of 180°
Central angletwo radiithe angle the arc subtends at the centre
02

The perpendicular from the centre

This is the theorem the chapter is built on, and it comes with a converse that is used just as often.

The perpendicular drawn from the centre of a circle to a chord bisects the chord. Conversely, the line joining the centre to the midpoint of a chord is perpendicular to that chord. The proof is a pair of congruent right-angled triangles sharing the perpendicular as a side, with two radii as hypotenuses.

d² + (c/2)² = r²the right-angled triangle formed by the radius, half the chord and the distance from the centre
r
the radius
c
the length of the chord
d
the distance from the centre to the chordmeasured along the perpendicular
Worked example

A chord of length 16 cm lies in a circle of radius 10 cm. How far is it from the centre?

  1. Drop the perpendicular from the centre O to the chord AB, meeting it at M.The perpendicular bisects the chord, so AM = MB = 8 cm.
  2. Triangle OMA is right-angled at M, with hypotenuse OA = 10 cm (a radius) and AM = 8 cm.The hypotenuse of this triangle is always a radius — that is what makes the method work.
  3. OM² = 10² − 8² = 100 − 64 = 36.Pythagoras, with the radius as the hypotenuse.
  4. OM = 6 cm.Sensible: 6 cm is less than the 10 cm radius, as any distance from the centre to a chord must be.

6 cm from the centre

Drag the chord longer and watch the distance to the centre shrink. When the chord becomes a diameter that distance reaches zero — which is exactly why the diameter is the longest chord in any circle.

03

Equal chords, equal arcs, equal distances

In the same circle, or in circles of equal radius, these three statements are all equivalent — each one implies the other two. Questions are set by giving you one and asking for another.

  • Equal chords are equidistant from the centre, and chords equidistant from the centre are equal.
  • Equal chords cut off equal arcs, and equal arcs are cut off by equal chords.
  • Equal chords subtend equal angles at the centre, and equal central angles stand on equal chords.
  • The longer the chord, the nearer it lies to the centre; the diameter, at zero distance, is the longest of all.

The condition "in the same circle" is not decoration

None of these results holds when the circles have different radii. A 10 cm chord in a circle of radius 6 cm behaves nothing like a 10 cm chord in a circle of radius 50 cm. If a question involves two circles, check that they are stated to be congruent before quoting any of these theorems.

04

Arc length and sector area

A sector of angle θ is the fraction θ/360 of the whole circle, so its arc and its area are the same fraction of the circumference and the total area. There is nothing else to remember.

The area of a segment is one step further: take the sector and subtract the triangle formed by the two radii and the chord.

arc length = (θ / 360) × 2πrsector area = (θ / 360) × πr²segment area = sector area − triangle areaθ in degrees; the triangle has area ½r² sin θ
Worked example

A sector of a circle of radius 7 cm has a central angle of 60°. Find its arc length and its area. Take π = 22/7.

  1. Fraction of the circle: 60/360 = 1/6.Write the fraction down first; both parts of the question use it.
  2. Circumference = 2 × (22/7) × 7 = 44 cm, so the arc is 44/6 = 7.33 cm.The 7 cancels neatly, which is why the examiner chose that radius.
  3. Total area = (22/7) × 49 = 154 cm².
  4. Sector area = 154/6 = 25.67 cm².Check the proportions: a sixth of the circle should have a sixth of the area, and it does.

Arc ≈ 7.33 cm; sector area ≈ 25.7 cm²

Before you leave this chapter

  1. A segment is cut off by a chord; a sector by two radii. They are not interchangeable words.
  2. The perpendicular from the centre bisects the chord — and the converse also holds.
  3. r² = d² + (c/2)² relates radius, distance from centre and chord length.
  4. In one circle: equal chords ⟺ equal arcs ⟺ equal central angles ⟺ equal distances from the centre.
  5. Arc and sector are the fraction θ/360 of the circumference and the area.

Practice questions

6 questions · 20 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

3 · 6 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
Define a chord and state which chord of a circle is the longest.
Model answer

A chord is a straight line segment whose two endpoints lie on the circle. The longest chord is the diameter, which passes through the centre.

Examiner tip. Naming the diameter and saying it passes through the centre are the two marks.

SQ2[2 marks]
State the theorem about the perpendicular drawn from the centre of a circle to a chord, and its converse.
Model answer

The perpendicular from the centre to a chord bisects the chord. Conversely, the line from the centre to the midpoint of a chord is perpendicular to it.

Examiner tip. Both directions are required. Many questions supply the midpoint and expect you to deduce the right angle, which is the converse rather than the theorem.

SQ3[2 marks]
Explain why two equal chords of the same circle are equidistant from its centre.
Model answer

The perpendicular from the centre bisects each chord, forming a right-angled triangle with the radius as hypotenuse and half the chord as one leg. Equal chords give equal halves, and the radii are equal, so by Pythagoras the third sides — the distances from the centre — must also be equal.

Examiner tip. Argue from r² = d² + (c/2)². If r and c match, d has no choice but to match too.

Solved numericals

2 · 8 marks

Full working, one step per line, with the marks shown where they are awarded.

N1[4 marks]
A chord of a circle is 24 cm long and lies 5 cm from the centre. Find the radius, and the length of a second chord that lies 12 cm from the centre.
Full working
  1. Half the chord is 12 cm; r² = 5² + 12² = 25 + 144 = 169perpendicular from the centre bisects the chord[1]
  2. r = 13 cm[1]
  3. For the second chord, (c/2)² = 13² − 12² = 169 − 144 = 25, so c/2 = 5same circle, so the same radius[1]
  4. Second chord = 10 cm[1]

r = 13 cm; the second chord is 10 cm long

Examiner tip. The second chord is further from the centre and comes out shorter, as it must. That check catches an inverted Pythagoras immediately.

N2[4 marks]
A sector has radius 14 cm and central angle 90°. Find its arc length, its area, and the perimeter of the sector. Take π = 22/7.
Full working
  1. Fraction = 90/360 = 1/4[1]
  2. Arc = ¼ × 2 × (22/7) × 14 = ¼ × 88 = 22 cm[1]
  3. Area = ¼ × (22/7) × 196 = ¼ × 616 = 154 cm²[1]
  4. Perimeter = arc + two radii = 22 + 14 + 14 = 50 cmthe two radii are part of the boundary and are often forgotten[1]

Arc 22 cm; area 154 cm²; perimeter 50 cm

Examiner tip. The perimeter of a sector is the arc plus both radii. Giving only the arc length is the standard wrong answer to this part.

Long questions

1 · 6 marks

Theory and numerical together, as they appear in the long-question section.

LQ1[6 marks]
In a circle of radius 10 cm, a chord AB subtends an angle of 90° at the centre O.
  1. Find the length of AB.
  2. Find the area of the minor sector OAB.
  3. Hence find the area of the minor segment cut off by AB. Take π = 3.14.
Mark scheme
  1. Triangle OAB is right-angled at O with OA = OB = 10 cmboth are radii[1]
  2. AB = √(100 + 100) = √200 = 14.1 cmaccept 10√2[1]
  3. Sector area = (90/360) × 3.14 × 100a quarter of the circle[1]
  4. = 78.5 cm²[1]
  5. Triangle area = ½ × 10 × 10 = 50 cm²the two radii are perpendicular, so they are the base and the height[1]
  6. Segment = 78.5 − 50 = 28.5 cm²segment = sector minus triangle[1]

(a) 14.1 cm (b) 78.5 cm² (c) 28.5 cm²

Examiner tip. A segment is always the sector minus the triangle. Draw the triangle inside the sector before calculating anything — the picture makes the subtraction obvious and stops you subtracting the wrong way round.