Mathematics notes, chapter 6

Angles and Tangents of Circles

Textbook Mathematics Form 3 (DLP), pages 128–167

What this chapter covers

Recognise the angle at the circumference and the central angle subtended by an arc, verify the relationships between them (same arc, arc length, twice the central angle, angle in a semicircle) and solve problems (6.1)
Recognise cyclic quadrilaterals, use the relationships between their angles (opposite angles add up to 180°, exterior angle equals the opposite interior angle) and solve problems (6.2)
Recognise tangents to circles and points of tangency, and verify that the radius is perpendicular to the tangent at the point of tangency (6.3)
Use the properties of two tangents from a point and the angle in the alternate segment, and solve problems involving tangents and common tangents (6.3)
Solve problems that combine angles in circles, cyclic quadrilaterals and tangents (6.4)

6.1Angle at the circumference and central angle subtended by an arc

Chord: a straight line joining two points on the circle. Arc: part of the circumference. The shorter arc is the minor arc, the longer one is the major arc. A chord subtends (cuts off) an arc, and an angle is "subtended by an arc" when its two arms end at the two ends of that arc.

Angle	Where its vertex is	Example in a circle with centre O
Angle at the circumference	On the circumference	∠PRQ, subtended by arc PQ
Central angle	At the centre O	∠POQ, subtended by the same arc PQ

A circle with chord PQ at the bottom and points A, B and C on the major arc; each of the three angles PAQ, PBQ and PCQ is 35 degrees — Angles at the circumference subtended by the same arc are equal. Here arc PQ is seen from three points and each angle is 35°.

The four rules

Rule	In words
1. Same arc	Angles at the circumference subtended by the same arc are equal.
2. Arc length	Angles at the circumference subtended by arcs of equal length are equal. The angle is proportional to the arc length: twice the arc, twice the angle.
3. Centre and circumference	The central angle is twice the angle at the circumference subtended by the same arc.
4. Diameter	The angle at the circumference subtended by a diameter is 90° (an angle in a semicircle).

Textbook summary: angles at the circumference subtended by arcs of the same length are equal, and the size of the angle is proportional to the arc length — Rule 2 as shown in the textbook.

Textbook figure: Mathematics Form 3 (DLP), p. 136.

Example 1 (textbook p. 139): central angle and angle at the circumference

Two circles with centre O and chord PQ. Left: R on the major arc, angle POQ 140 degrees, angle PRQ is x. Right: R on the minor arc, angle PRQ is y

In both circles the central angle ∠POQ = 140°.
Left, R is on the major arc: x = ½ × 140° = 70°.
Right, R is on the minor arc, so it sees the major arc, whose central angle is 360° − 140° = 220°: y = ½ × 220° = 110°.
Check: x + y = 180°.

Textbook: four circles showing the central angle 2x and the angle at the circumference x subtended by the same arc — Rule 3: the central angle is twice the angle at the circumference, wherever the point is on the circle.

Textbook figure: Mathematics Form 3 (DLP), p. 139.

Halving the central angle only works when the point on the circumference is on the major arc side. If the point is on the minor arc, it sees the other (reflex) central angle, 360° − ∠POQ, and the angle at the circumference is half of that, so it is more than 90°.

Example 2 (textbook p. 136): arc length and angle

A circle where arc PQ of 4 cm subtends 20 degrees at A and arc RS of 10 cm subtends x at B

The angle is proportional to the arc length, so the ratio of angles equals the ratio of arcs.
x ÷ 20° = 10 cm ÷ 4 cm
x = 20° × 10 ÷ 4 = 50°

Central angles follow the same two rules: equal arcs give equal central angles, and the central angle is proportional to the arc length. So 4 cm of arc is 40° at the centre and 20° at the circumference, and a whole circle of 360° is 36 cm.

Example 3 (textbook p. 142): angle in a semicircle

A circle with diameter PQ, centre O, and R on the circumference; angle RPQ is 28 degrees, angle PRQ is a right angle and angle PQR is y

PQ is a diameter, so ∠PRQ = 90°.
Angles in triangle PQR add up to 180°: y = 180° − 90° − 28° = 62°.

Textbook rule: the angle at the circumference subtended by a diameter is 90 degrees

Textbook figure: Mathematics Form 3 (DLP), p. 142.

(a) Arc AB subtends 22° at the circumference. Arc CD is three times as long as arc AB. What angle does arc CD subtend at the circumference?
(b) In a circle with centre O, ∠POQ = 130°. Find ∠PRQ when R is on the major arc, and when R is on the minor arc.
(c) AB is a diameter and C is on the circle. ∠CAB = 40°. Find ∠ABC.

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(a) 3 × 22° = 66°
(b) Major arc: ½ × 130° = 65°. Minor arc: ½ × (360° − 130°) = 115°
(c) ∠ACB = 90°, so ∠ABC = 180° − 90° − 40° = 50°

6.2Cyclic quadrilaterals

A cyclic quadrilateral is a four-sided figure whose four vertices all lie on the circumference of a circle. If even one vertex is inside or outside the circle, it is not cyclic. Opposite angles are the two angles that face each other, such as ∠A and ∠C.

Textbook: cyclic quadrilateral with angles x, y, p, q; the sum of opposite interior angles is 180 degrees — The sum of opposite interior angles of a cyclic quadrilateral is 180°.

Textbook figure: Mathematics Form 3 (DLP), p. 145.

Example 4 (textbook p. 146): opposite angles

Cyclic quadrilateral ABCD with angle ABC 80 degrees, angle DAB 85 degrees, angle ADC x and angle BCD y

x is opposite ∠ABC: x = 180° − 80° = 100°.
y is opposite ∠DAB: y = 180° − 85° = 95°.
Check: the four angles add up to 80° + 100° + 85° + 95° = 360°.

Exterior angle of a cyclic quadrilateral

Extend one side of a cyclic quadrilateral. The angle formed outside is the exterior angle. It is equal to the opposite interior angle (the interior angle at the vertex facing it).

Example 5 (textbook p. 147): exterior angle

Cyclic quadrilateral PQRS with QR extended to T; the exterior angle SRT is 95 degrees and the opposite interior angle SPQ is y

Exterior ∠SRT sits on a straight line with ∠QRS, so ∠QRS = 180° − 95° = 85°.
∠SPQ is opposite ∠QRS, so y = 180° − 85° = 95°.
So the exterior angle equals the opposite interior angle: y = 95°. Use this shortcut to skip a step.

Name opposite angles by going across the quadrilateral: in PQRS, ∠P and ∠R are opposite, and so are ∠Q and ∠S.

Adding the two adjacent angles (next to each other) to get 180°. Only opposite angles add to 180°. Adjacent angles in a cyclic quadrilateral have no fixed sum.

(a) In cyclic quadrilateral PQRS, ∠PQR = 72°. Find ∠PSR.
(b) A side of a cyclic quadrilateral is extended and the exterior angle is 64°. What is the opposite interior angle?

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(a) ∠PSR = 180° − 72° = 108°
(b) 64° (exterior angle = opposite interior angle)

6.3Tangents to circles

A tangent to a circle is a straight line that touches the circle at only one point. That point is the point of tangency. A line that cuts the circle at two points is not a tangent.

Tangent and radius

Textbook: the radius to the point of tangency makes a 90 degree angle with the tangent

Textbook figure: Mathematics Form 3 (DLP), p. 152.

The radius drawn to the point of tangency is perpendicular to the tangent: the angle between them is 90°.

Example 6 (textbook p. 152): tangent and radius

Line ABC touches the circle with centre O at B; angle AOB is 55 degrees and angle OAB is x

OB is a radius to the point of tangency, so ∠OBA = 90°.
In triangle OAB: x = 180° − 90° − 55° = 35°.

Two tangents from one point

Textbook: two tangents from A touch the circle at B and C; BA equals CA and angles BOA and COA are equal

Textbook figure: Mathematics Form 3 (DLP), p. 154.

If tangents from a point P touch a circle with centre O at A and B, then:

the two tangent lengths are equal: PA = PB
∠AOP = ∠BOP and ∠OPA = ∠OPB (OP bisects the angles)
∠OAP = ∠OBP = 90°, so OAPB is a kite

Example 7 (textbook p. 154): two tangents

Tangents PA and PB to a circle with centre O; angle AOB is 110 degrees and angle APB is y

In the kite OAPB, the angles at A and B are 90°.
y = 360° − 90° − 90° − 110° = 70°
OP bisects ∠APB, so ∠OPA = 35°.

In the right-angled triangle OAP, use Pythagoras or trigonometry: OP² = OA² + AP². For example OA = 5 cm and OP = 13 cm give AP = √(13² − 5²) = 12 cm.

Angle in the alternate segment

The angle between a tangent and a chord at the point of tangency is equal to the angle in the alternate segment, that is the angle at the circumference on the opposite side of the chord, subtended by the same chord.

Textbook: tangent PQR with chords QS, ST, TQ; angle x equals angle y and angle theta equals angle beta

Textbook figure: Mathematics Form 3 (DLP), p. 155.

Example 8 (textbook p. 156): alternate segment

Tangent PQR with chords QS and QT and triangle QST; angle PQS is 55 degrees, angle RQT is 65 degrees, angle QTS is x and angle QST is y

x = ∠QTS is in the alternate segment of chord QS, so x = ∠PQS = 55°.
y = ∠QST is in the alternate segment of chord QT, so y = ∠RQT = 65°.
Check with the triangle: ∠SQT = 180° − 55° − 65° = 60°, and 55° + 60° + 65° = 180° on the straight line PQR.

Using the angle on the same side of the chord. The matching angle is always across the chord, in the alternate segment. A common slip is to match ∠PQS with ∠QST, which belongs to the other chord, QT.

Common tangents to two circles

A common tangent touches both circles. How many there are depends on how the circles sit.

Position of the two circles	Common tangents
Not touching each other	4
Touching at one point	3
Intersecting (cutting each other at two points)	2
Overlapping (one circle inside the other, touching at one point)	1

Textbook: pairs of circles not touching, touching, intersecting and overlapping, with their common tangents drawn

Textbook figure: Mathematics Form 3 (DLP), p. 157.

Example 9 (textbook p. 158): length of a common tangent

Circles with centres A and B and radii 8 cm and 3 cm, 13 cm apart, with common tangent PQ; a dashed line BX parallel to PQ meets AP at X

Draw BX parallel to PQ, with X on AP. Then PQXB is a rectangle: PX = QB = 3 cm and BX = PQ.
AX = AP − PX = 8 − 3 = 5 cm.
In right-angled triangle AXB: BX² = AB² − AX² = 13² − 5² = 144, so PQ = BX = 12 cm.

(a) PA and PB are tangents to a circle with centre O. ∠APB = 50°. Find ∠AOB.
(b) The radius of a circle is 7 cm and a point P is 25 cm from the centre. Find the length of the tangent from P.
(c) A tangent and a chord make an angle of 38° at the point of tangency. What is the angle in the alternate segment?

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(a) ∠AOB = 360° − 90° − 90° − 50° = 130°
(b) √(25² − 7²) = √576 = 24 cm
(c) 38°

6.4Angles and tangents of circles

Many problems use two or three of the rules. Draw the diagram, mark what you know, and give a reason for each step.

What you see	Rule to use
Two angles at the circumference on the same arc	Equal
An angle at the centre and one at the circumference on the same arc	Central angle = 2 × angle at the circumference
A diameter	Angle at the circumference = 90°
Four vertices on a circle	Opposite angles add to 180°; exterior = opposite interior
A tangent and a radius	Angle = 90°
Two tangents from one point	Equal lengths; kite OAPB
A tangent and a chord	Angle = angle in the alternate segment
Radii OA and OB	Isosceles triangle OAB: equal base angles

Example 10 (textbook p. 160): combined problem

Circle with centre O; tangent PAQ at A; chord AB, angle PAB 50 degrees; C on the circle; x is angle ACB, y is angle AOB, z is angle OAB

(a) x = ∠ACB is in the alternate segment of chord AB, so x = ∠PAB = 50°.
(b) y = ∠AOB is the central angle on the same arc AB, so y = 2 × 50° = 100°.
(c) OA = OB (radii), so triangle OAB is isosceles: z = (180° − 100°) ÷ 2 = 40°.
Check: OA is a radius to the point of tangency, so ∠OAP = 90° and z = 90° − 50° = 40°. Both routes agree.

When two different routes give the same answer you can be confident. In the example above, the isosceles triangle and the tangent-radius right angle both give 40°.

PA is a tangent at A to a circle with centre O. B is on the circle and ∠PAB = 64°. Find ∠AOB and ∠OAB.

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Angle in the alternate segment = 64°, so ∠AOB = 2 × 64° = 128°.
∠OAB = 90° − 64° = 26° (check: (180° − 128°) ÷ 2 = 26°)