Mathematics notes, chapter 5

Circles

Textbook Mathematics Form 2 (DLP), pages 74–97

What this chapter covers

Name the parts of a circle (centre, radius, diameter, chord, arc, sector, segment, circumference) and explain the properties of a circle (5.1)
Construct a circle, a diameter through a point, a chord of given length and a sector of given angle using compasses, ruler and protractor (5.1)
Explain that a diameter is an axis of symmetry, that a radius perpendicular to a chord bisects it, that perpendicular bisectors of chords meet at the centre, and that equal chords give equal arcs and are equidistant from the centre (5.2)
Find the centre and radius of a circle by construction, and solve problems on the symmetrical properties of chords using Pythagoras' theorem (5.2)
Define π as circumference ÷ diameter and use the circumference formulae C = πd = 2πr (5.3)
Derive and use the area formula A = πr² (5.3)
Find the length of an arc and the area of a sector, and work backwards to the radius or the angle at the centre (5.3)
Solve problems involving circles, including combinations of shapes and the area of an annulus (5.3)

Wheels, coins, clock faces, round tables and satellite dishes are all circles. A circle is the path of a point that always stays the same distance from one fixed point. In this chapter you will name the parts of a circle, use its symmetry to find unknown lengths of chords, and calculate its circumference, area, arcs and sectors.

English	Bahasa Melayu
circle, circumference	bulatan, lilitan
centre, radius, diameter	pusat, jejari, diameter
chord, arc	perentas, lengkok
sector, segment	sektor, tembereng
minor, major	minor, major
symmetry, axis of symmetry	simetri, paksi simetri
perpendicular bisector	pembahagi dua sama serenjang
area of a circle	luas bulatan

5.1Properties of circles

Parts of a circle

A circle is the path of a point that moves so that it is always the same distance from a fixed point, the centre. The distance is the radius, and the length of the curve all the way round is the circumference.

A circle with its centre O, a radius, a horizontal diameter through O, a chord below the diameter, an arc on the right and the circumference — Lines and curves of a circle (textbook p. 76)

Part	What it is
Centre	The fixed point; every point on the circumference is the same distance from it
Radius	A straight line from the centre to any point on the circumference
Diameter	A chord that passes through the centre (diameter = 2 × radius)
Chord	A straight line joining any two points on the circumference
Arc	Part of the circumference (a minor arc is shorter than a major arc)
Sector	The region enclosed by two radii and an arc
Segment	The region enclosed by a chord and an arc

Left: a circle with a small sector shaded blue as the minor sector and the rest as the major sector. Right: a circle with a chord near the top; the small part above is the minor segment and the rest is the major segment — Minor and major sectors, minor and major segments

The diameter is the longest chord of a circle. A radius is not a chord, because one end is the centre, not a point on the circumference.

Mixing up a sector and a segment. A sector has two radii (it looks like a slice of pizza). A segment is cut off by one chord and has no centre corner.

Example 1 (textbook p. 77)

Using the first figure above, name (a) the longest chord, (b) a line from the centre to the circumference, (c) the curved line that is one part of the circumference.

(a) Diameter (b) Radius (c) Arc

Name (a) a straight line that joins two points on the circumference but does not pass through the centre, (b) the region between a chord and its arc, (c) the region between two radii and an arc.

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(a) Chord
(b) Segment
(c) Sector

Constructing a circle and its parts

Use a pair of compasses, a ruler and a protractor.

Circle of given radius: mark the centre O, open the compasses to the radius on a ruler, put the sharp point on O and draw the circle.
Diameter through a point Q on the circle: join Q to the centre O with a ruler and extend the line until it meets the circumference on the other side.
Chord of given length from a point P: open the compasses to the length, put the sharp point on P and draw an arc that cuts the circle at A. Join P to A. (The chord can be at most one diameter long.)
Sector with a given angle: draw the circle and a radius OA. Use the protractor at O to mark the angle, draw the second radius OB. Sector AOB is the region between them.

Three steps: draw radius OA of 2 cm; mark an angle of 60 degrees at O with a protractor; draw radius OB and the arc to give the sector AOB — Constructing a sector of 60° with radius 2 cm (textbook p. 78)

To read an angle with a protractor, put its centre on the vertex of the angle and line up the 0° line with one arm. Then read from the scale that starts at 0° on that arm.

Describe how to draw a chord of length 4 cm from a point P on a circle of radius 3 cm.

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Open the compasses to 4 cm. Put the sharp point on P and draw an arc that cuts the circumference at A. Join P and A with a ruler: PA is the chord (4 cm is less than the diameter of 6 cm, so it is possible).

5.2Symmetrical properties of chords

Features in a circle

A circle looks the same after it is folded along any diameter, so a circle has infinitely many axes of symmetry. This one idea gives all the properties below.

1. A diameter is an axis of symmetry of the circle.
2. A radius that is perpendicular to a chord bisects the chord (and a radius that bisects a chord is perpendicular to it).
3. The perpendicular bisectors of two chords meet at the centre of the circle.
4. Chords that are equal in length cut off arcs of the same length (and the other way round).
5. Equal chords are the same distance from the centre (and the other way round).

A circle with centre O and two equal parallel chords AB and CD marked with double ticks; OP and OQ are perpendicular to the chords and are equal — AB = CD, so OP = OQ and arcs AB and CD are equal (textbook pp. 82–83)

"Distance from the centre" always means the perpendicular distance, the shortest distance from O to the chord.

Two radii and a chord form an isosceles triangle, so the angles opposite the equal radii are equal. The perpendicular from the centre to the chord also splits that triangle into two right-angled triangles, where you can use Pythagoras' theorem: (half the chord)² + (distance from centre)² = radius².

Example 2 (textbook p. 83)

O is the centre of the circle and MN is a chord. K is the midpoint of MN and OK = 3 cm, NK = 4 cm. Find (a) ON, (b) an angle equal to ∠ONK.

Circle with centre O, chord MN, K the midpoint of MN with a right angle at K, OK = 3 cm and NK = 4 cm

OK ⊥ MN, so triangle OKN is right-angled at K.
ON² = OK² + NK² = 3² + 4² = 9 + 16 = 25, so ON = 5 cm.
OM = ON (radii), so triangle OMN is isosceles and ∠ONK = ∠OMK.

Example 3 (textbook p. 83)

Radius OP is perpendicular to chord MN at S. The radius is 10 cm and OS = 8 cm. Find MN.

Circle with centre O, radius OP perpendicular to chord MN at S; OP = 10 cm, OS = 8 cm; MS and SN marked equal

OM is also a radius, so OM = 10 cm.
MS² = OM² − OS² = 10² − 8² = 100 − 64 = 36, so MS = 6 cm.
MS = SN (a radius perpendicular to a chord bisects it), so MN = 2 × 6 = 12 cm.

Giving MS (half the chord) as the answer, or using the hypotenuse the wrong way round (adding the squares to find a short side). The radius is always the hypotenuse of the right-angled triangle.

Centre and radius of a circle

To find the centre of a circle drawn on paper, draw two chords, for example PQ and PR, and construct the perpendicular bisector of each. They meet at the centre O. The distance from O to the circumference is the radius.

A circle with chords PQ and PR drawn from P; the perpendicular bisector of each chord is a dashed line; the dashed lines cross at the centre O — The perpendicular bisectors of two chords meet at the centre (textbook p. 84)

Solving problems using the symmetry of chords

Example 4 (textbook p. 84)

RS and TU are equal chords. POQ is a straight line through the centre O. OP = 5 cm and RS = 24 cm. Calculate (a) PR, (b) the radius.

Circle with centre O; equal chords RS above O and TU below O; the straight line POQ is perpendicular to both; OP = 5 cm and RS = 24 cm

OP ⊥ RS, so P is the midpoint: PR = 24 ÷ 2 = 12 cm.
Radius OR² = PR² + OP² = 12² + 5² = 144 + 25 = 169, so OR = 13 cm.
(Equal chords are the same distance from O, so OQ = 5 cm as well, and arcs RS and TU are equal.)

Example 5 (textbook p. 85)

A round window has a diameter of 50 cm. Two parallel iron rods US = 48 cm and PR are on opposite sides of the centre. The distance QT between them is 31 cm. Find the length of PR.

Circle of radius 25 cm with centre O; chord US below O of length 48 cm; chord PR above O; QT is perpendicular to both chords and equals 31 cm

Radius = 50 ÷ 2 = 25 cm. Half of US = 24 cm.
OT = √(25² − 24²) = √(625 − 576) = √49 = 7 cm.
OQ = QT − OT = 31 − 7 = 24 cm.
PQ = √(25² − 24²) = √49 = 7 cm, so PR = 2 × 7 = 14 cm.

Two parallel chords on opposite sides of O: distance between them = sum of the two distances from O. On the same side: the difference.

A circle has radius 13 cm. A chord is 12 cm from the centre. Find the length of the chord.

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Half the chord = √(13² − 12²) = √(169 − 144) = √25 = 5 cm
Chord = 2 × 5 = 10 cm

5.3Circumference and area of a circle

Circumference and diameter: the number π

If you measure the circumference and the diameter of any circular object (a pail, a tyre, a stopwatch) and divide, you always get about 3.14.

A circle with its diameter d. Below, a line made of three lengths d and a small extra piece of 0.14 d, which together equal the circumference — The circumference is a little more than three diameters (textbook p. 86)

The ratio circumference ÷ diameter is the same for every circle. It is called π (pi). Use π = 227 or π = 3.142 as the question tells you.
Circumference = πd = 2πr

Area of a circle

Cut a circle into many thin sectors and arrange them side by side. The shape becomes nearly a rectangle with height r and base = half the circumference = πr.

A circle cut into eight sectors with radius r marked, and a rectangle of base pi r and height r with area pi r times r equals pi r squared — Deriving the area of a circle (textbook p. 87)

Area of a circle = πr². The radius is squared, not the diameter and not π.

Using the diameter in πr², or working out (πr)² instead of π × r². If you are given the diameter, halve it first.

Circumference and area problems

You know	You want	Method
radius or diameter	circumference	C = 2πr or πd
circumference	radius	r = C ÷ 2π, then area if needed
area	radius	r² = A ÷ π, then take the square root

Example 6 (textbook p. 88)

Calculate the circumference of a circle with (a) diameter 14 cm (π = 227), (b) radius 21.3 cm (π = 3.142).

(a) C = πd = 227 × 14 = 44 cm

(b) C = 2πr = 2 × 3.142 × 21.3 = 133.85 cm (2 d.p.)

Example 7 (textbook p. 88)

The circumference of a circle is 88 cm. Find its diameter (π = 227).

πd = 88, so 227 × d = 88 and d = 88 × 722 = 28 cm

Example 9 (textbook p. 89)

The area of a circle is 616 cm². Find its radius and diameter (π = 227).

πr² = 616, so r² = 616 × 722 = 196 and r = √196 = 14 cm. Diameter = 2 × 14 = 28 cm.

Example 10 (textbook p. 89)

The circumference is 66 cm. Find the area (π = 227).

2πr = 66, so 2 × 227 × r = 66 and r = 66 × 744 = 10.5 cm
Area = πr² = 227 × 10.5² = 346.5 cm²

Parts of a circle: a semicircle is ½ of the circle, a quadrant is ¼. For a quadrant of radius 7 cm (π = 227): area = ¼ × 227 × 7² = 38.5 cm²; for a semicircle: ½ × 227 × 7² = 77 cm².

(π = 227) Find (a) the circumference of a circle of radius 14 cm, (b) the area of a circle of diameter 14 cm.

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(a) 2 × 227 × 14 = 88 cm
(b) r = 7 cm, area = 227 × 7² = 154 cm²

Length of an arc and area of a sector

An arc is part of the circumference, and a sector is part of the circle. Each is in the same proportion to the whole as its angle at the centre is to 360°.

A circle with centre O, a sector AOB with an angle of 60 degrees at O and the arc AB drawn in red — Arc AB and sector AOB (textbook p. 90)

Length of arc = θ360° × 2πr
Area of sector = θ360° × πr²
where θ is the angle at the centre.

Remember what each fraction does: θ360° picks the fraction of a whole turn. For a major arc or sector, use 360° − θ.

Example 12 (textbook p. 90)

Radius 14 cm, centre O. Calculate the length of minor arc PQ with a 60° angle at the centre. Give the answer to 2 d.p. (π = 227)

Circle centre O radius 14 cm with angle POQ equal to 60 degrees and the minor arc PQ in red

Length = 60°360° × 2 × 227 × 14 = 16 × 88 = 14.67 cm

Example 13 (textbook p. 90)

Radius 21 cm and ∠ROS = 72°. Calculate the length of the major arc RS. (π = 227)

Circle centre O radius 21 cm with angle ROS equal to 72 degrees; the major arc RS is drawn in red

Angle at the centre for the major arc = 360° − 72° = 288°
Length = 288°360° × 2 × 227 × 21 = 105.6 cm

Example 14 (textbook p. 91)

An arc is 11 cm long and its angle at the centre is 45°. Find the radius. (π = 227)

45°360° × 2 × 227 × r = 11, so 18 × 447 × r = 11, 1114 r = 11 and r = 14 cm

Example 15 (textbook p. 91)

Radius 21 mm, ∠MON = 100°. Calculate the area of the minor sector MON. (π = 227)

Circle centre O radius 21 mm with the minor sector MON, angle 100 degrees, shaded

Area = 100°360° × 227 × 21² = 385 mm²

Example 16 (textbook p. 92)

The area of sector QOP is 18.48 cm² and the radius is 12 cm. Find the angle θ at the centre. (π = 227)

θ360° × 227 × 12² = 18.48, so θ = 18.48 × 360° ÷ (227 × 144) = 14.7°

Writing the angle of a major arc as the small angle. If the question asks for the major arc or sector, use 360° − θ.

(a) Find the length of an arc of a circle of radius 6 cm with an angle of 150° at the centre. (π = 3.142) (b) Find the area of a sector of angle 135° and radius 8 cm. (π = 227)

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(a) 150°360° × 2 × 3.142 × 6 = 15.71 cm
(b) 135°360° × 227 × 8² = 75.43 cm²

Solving problems involving circles

Most problems combine shapes. Break the figure into rectangles, triangles and parts of circles, find each area (or length), then add or subtract.

Example 17 (textbook p. 92)

A park is a rectangle 63 m by 58 m. A quadrant of radius 7 m is planted with flowers at each corner. A circular fish pond of diameter 28 m is in the middle. The rest is grass. Find the area of grass. (π = 227)

A rectangle 63 m by 58 m with a pink quarter circle of radius 7 m in each corner and a blue circular pond of diameter 28 m in the middle

Park = 63 × 58 = 3 654 m²
Four quadrants make one whole circle of radius 7 m: 227 × 7² = 154 m²
Pond: radius 14 m, area = 227 × 14² = 616 m²
Grass = 3 654 − 154 − 616 = 2 884 m²

An annulus

The region between two circles with the same centre is an annulus. Its area is the big circle minus the small circle.

Two circles with the same centre O; the inner circle has radius 6 cm and the outer circle has radius 12 cm; the ring between is shaded — An annulus (textbook p. 92)

Area of annulus = πR² − πr² = π(R² − r²), where R is the outer radius and r is the inner radius.

With R = 12 cm and r = 6 cm: area = π(144 − 36) = 108π cm² (about 339.43 cm² using 227).

Working out π(R − r)². The squares must be done first and then subtracted: R² − r², not (R − r)².

Find the area of an annulus with outer radius 10 cm and inner radius 6 cm, in terms of π and then using π = 227 (2 d.p.).

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π(10² − 6²) = π(100 − 36) = 64π cm²
64 × 227 = 201.14 cm²

Summary

Idea	Key fact	Example
Parts	Diameter is the longest chord; sector has two radii and an arc; segment has a chord and an arc	Diameter = 2 × radius
Symmetry	A radius perpendicular to a chord bisects it; equal chords are equidistant from the centre and cut equal arcs	r = 10, OS = 8: MN = 12
Chord problems	Right-angled triangle: r² = (half chord)² + (distance)²	5² + 12² = 13²
Circumference	C = πd = 2πr	r = 21.3: 133.85 cm
Area	A = πr²	r = 14: 616 cm²
Arc and sector	θ360° of 2πr and of πr²	60°, r = 14: 14.67 cm
Annulus	π(R² − r²)	R = 12, r = 6: 108π