Circle geometry - Wiley: · PDF file812 Maths Quest 10 + 10A measurement anD geometry The angle subtended at the centre of a circle is twice the angle subtended at the circumference,

measurement anD geometrymeasurement anD geometry

Circle geometry19.1 OverviewWhy learn this?For thousands of years humans have been fascinated by circles. Since they fi rst looked upwards towards the sun and moon, which, from a distance at least, looked circular, humans have created circular monuments to nature. The most famous circular invention, one that has been credited as the most important invention of all, is the wheel. Scholars as early as Socrates and Plato have been fascinated with the sheer beauty of the properties of circles, and many scholars made a life’s work out of studying them. Euclid was probably the most famous of these. It is in circle geometry that the concepts of congruence and similarity, studied earlier, have a powerful context.

What do you know? 1 tHInK List what you know about circle geometry. Use

a thinking tool such as a concept map to show your list.2 PaIr Share what you know with a partner and then with

a small group.3 sHare As a class, create a thinking tool such as a large concept

map that shows your class’s knowledge of circle geometry.

Learning sequence19.1 Overview19.2 Angles in a circle19.3 Intersecting chords, secants and tangents19.4 Cyclic quadrilaterals19.5 Tangents, secants and chords19.6 Review ONLINE ONLY

toPIC 19

measurement anD geometry

c19CircleGeometry.indd 808 08/07/16 1:27 PM

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WatCH tHIs VIDeoThe story of mathematics:Piscopia

searchlight ID: eles-2022


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810 Maths Quest 10 + 10A


19.2 Angles in a circleIntroduction • A circle is a set of points that lie a fi xed distance (the radius) from a fi xed point (the

centre). • In circle geometry, there are many theorems that can be used to solve problems. It is

important that we are also able to prove these theorems. • To prove a theorem:

1. state the aim of the proof2. use given information and previously established theorems to establish the result3. give a reason for each step of the proof4. state a clear conclusion.

Parts of a circlePart (name) Description Diagram

Centre The middle point, equidistant from all points on the circumference. It is usually shown by a dot and labelled O.

O

Circumference The outside length or the boundary forming the circle. It is the circle’s perimeter.

O

Radius A straight line from the centre to any point on the circumference

O

Diameter A straight line from one point on the circumference to another, passing through the centre

O

Chord A straight line from one point on the circumference to another O

Segment The area of the circle between a chord and the circumference. The smaller segment is called the minor segment and the larger segment is the major segment.

O

O

int-2795int-2795int-2795


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Topic 19 • Circle geometry 811


Part (name) Description Diagram

Sector An area of a circle enclosed by 2 radii and the circumference O

O

Arc A portion of the circumference

O

O

Tangent A straight line that touches the circumference at one point only

O

Secant A chord extended beyond the circumference on one side O

Angles in a circle • In the diagram at right, chords AC and BC form the angle ACB.

Arc AB has subtended angle ACB.

• Theorem 1 Code The angle subtended at the centre of a circle is twice the angle subtended at the circumference, standing on the same arc.

Proof:Let ∠PRO = x and ∠QRO = y

RO = PO = QO (radii of the same circle are equal)

∠RPO = x and ∠RQO = y

∠POM = 2x (exterior angle of triangle) and ∠QOM = 2y (exterior angle of triangle)

∠POQ = 2x + 2y= 2(x + y)

which is twice the size of ∠PRQ = x + y.

A B

C

x y

O

R

P

Q

M

P

R

OQ


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The angle subtended at the centre of a circle is twice the angle subtended at the circumference, standing on the same arc.

• Theorem 2 Code All angles that have their vertex on the circumference and are subtended by the same arc are equal.Proof:Join P and Q to O, the centre of the circle.Let∠PSQ = x

∠POQ = 2x (angle at the centre is twice) the angle at the circumference)

∠PRQ = x (angle at the circumference is half the angle of the centre)∠PSQ = ∠PRQ.

Angles at the circumference subtended by the same arc are equal.The application of the first two circle geometry theorems can be seen in the following

worked example.

Find the values of the pronumerals in the diagram at right, giving reasons for your answers.

tHInK WrIte

1 Angles x and 46° are angles subtended by the same arc and both have their vertex on the circumference.

x = 46°

2 Angles y and 46° stand on the same arc. The 46° angle has its vertex on the circumference and y has its vertex at the centre. The angle at the centre is twice the angle at the circumference.

y = 2 × 46°= 92°

y

x46°O

WorKeD eXamPLe 1

• Theorem 3 Code Angles subtended by the diameter, that is, angles in a semicircle, are right angles.

In the diagram at right, PQ is the diameter. Angles a, b and c are right angles. This theorem is in fact a special case of Theorem 1.Proof:∠POQ = 180° (straight line)

Let S refer to the angle at the circumference subtended by the diameter. In the figure, S could be at the points where a, b and c are represented on the diagram.∠PSQ = 90° (angle at the circumference is half the angle at the centre)Angles subtended by a diameter are right angles.

Constructing a tangentThere are a number of ways to construct a tangent to a circle, as explained using the following steps.

O

R

P

S

QP Q

OP

c b

a

Q


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1. Drawacircleofradius5 cmandcentreO.2. Draw a radius.3. Call the point of intersection of the radius and the circumference, P.4. Extend this radius through P to the point Q, 5 cm outside the circle.5. Using O and Q as centres, draw intersecting arcs above and below the line OQ.6. Draw a straight line joining the points of intersection. This line is the tangent.7. What do you notice about the angle between OQ and the tangent?8. Investigate another technique for constructing a tangent to a circle.9. Write a set of instructions for this method of constructing a tangent.

O QP

• Theorem 4 Code If a radius is drawn to any point on the circumference and a tangent is drawn at the same point, then the radius will be perpendicular to the tangent.

In the diagram at right, the radius is drawn to a point, P, on the circumference. The tangent to the circle is also drawn at P. The radius and the tangent meet at right angles, that is, the angle at P equals 90°.

Find the values of the pronumerals in the diagram at right, giving a reason for your answer.

tHInK WrIte

1 Angle z is subtended by the diameter. Use an appropriate theorem to state the value of z.

z = 90°

2 Angle s is formed by a tangent and a radius, drawn to the point of contact. Apply the corresponding theorem to find the value of s.

s = 90°

O

zs

WorKeD eXamPLe 2

• Theorem 5 Code The angle formed by two tangents meeting at an external point is bisected by a straight line joining the centre of the circle to that external point.Proof:

S

R

T

O

PO


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Consider ΔSOR and ΔSOT.OR = OT (radii of the same circle are equal)OS is common.∠ORS = ∠OTS = 90° (angle between a tangent and radii is 90°)∴ ΔSOR ≅ ΔSOT (RHS)So ∠ROS = ∠TOS and ∠OSR = ∠OST ( corresponding angles in congruent triangles

are equal).The angle formed by two tangents meeting at an external point is bisected by a straight line joining the centre of the circle to the external point.

Given that BA and BC are tangents to the circle, find the values of the pronumerals in the diagram at right. Give reasons for your answers.

tHInK WrIte

1 Angles r and s are angles formed by the tangent and the radius, drawn to the same point on the circle. State their size.

s = r = 90°

2 In the triangle ABO, two angles are already known and so angle t can be found using our knowledge of the sum of the angles in a triangle.

ΔABO: t + 90° + 68° = 180°t + 158° = 180°

t = 22°

3 ∠ABC is formed by the two tangents, so the line BO, joining the vertex B with the centre of the circle, bisects this angle. This means that angles t and u are equal.

∠ABO = ∠CBO∠ABO = t = 22°, ∠CBO = u u = 22°

4 ΔAOB and ΔCOB are similar triangles.

In ΔAOB and ΔCOBr + t + 68° = 180° s + u + q = 180° r = s = 90° (proved previously) t = u = 22° (proved previously) ∴ q = 68°

B

A

C

O tu

r

q

s

68°

WorKeD eXamPLe 3


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Exercise 19.2 Angles in a circle InDIVIDuaL PatHWays

⬛PraCtIseQuestions:1a–e, 2a–c, 3a–d, 4–7

⬛ConsoLIDateQuestions:1d–i, 2b–f, 3c–f, 4–10

⬛masterQuestions:1–11

FLuenCy

1 WE1 Find the values of the pronumerals in each of the following, giving reasons for your answers.a

A B

x30° b

yx

P Q

25°

c

x

S

R

32°

d x

y40°

e

x

30°•O

A B

f

•

x80°

A

B

O

g

42°O•x

A

B

h

x50°

y

O•

i

x 28°

O•A

B

2 WE2 Find the values of the pronumerals in each of the following fi gures, giving reasons for your answers.a

r

s•

b

••

u tc

•

m n

d

• x38°

e

x

•O

f

x

y

•O75°

unDerstanDIng

3 WE3 Given that AB and DB are tangents, fi nd the value of the pronumerals in each of the following, giving reasons for your answers.a

x

yw

z

O 70°•

A

B

D

b

t

s

r

O40° •

A

B

D

reFLeCtIon What are the common steps in proving a theorem?

doc-5390doc-5390doc-5390doc-5390doc-5390doc-5390

doc-5391doc-5391doc-5391





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c

xyzO 20°•

A

B

D

d

xs y

zr

O

70°

•

A B

D

e

y zx

O

20°

•

A B

D f

xy

15°

A

D

B

•

Oz

4 MC Note: There may be more than one correct answer.In the diagram at right, which angle is subtended by the same arc as ∠APB?a ∠APCB ∠BPCC ∠ABPD ∠ADB

5 MC Note: There may be more than one correct answer.Referring to the diagram at right, which of the statements is true?a 2∠AOD = ∠ABDB ∠AOD = 2∠ACD

C ∠ABF = ∠ABDD ∠ABD = ∠ACD

reasonIng

6 Values are suggested for the pronumerals in the diagram at right. AB is a tangent to a circle and O is the centre. In each case give reasons to justify suggested values.a s = t = 45° b r = 45°c u = 65° d m = 25°

e n = 45°

7 Set out below is the proof of this result: The angle at the centre of a circle is twice the angle at the circumference standing on the same arc.

P Q

R

a

bO

A

C B

P

D

A

CB

F

O

D

Ar s

tu

n

mC

B

F

O

D

25°


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Copy and complete the following to show that ∠POQ = 2 × ∠PRQ.Construct a diameter through R. Let the opposite end of the diameter be S.Let ∠ORP = x and ∠ORQ = y.OR = OP ( )∠OPR = x ( )∠SOP = 2x (exterior angle equals )OR = OQ ( )∠OQR = ( )∠SOQ = ( )Now ∠PRQ = and ∠POQ = .Therefore ∠POQ = 2 × ∠PRQ.

8 Prove that the segments formed by drawing tangents from an external point to a circle are equal in length.

9 Use the fi gure drawn below to prove that angles subtended by the same arc are equal.

P

SR

O

Q

ProBLem soLVIng

10 Use your knowledge of types of triangles, angles in triangles and the fact that the radius of a circle meets the tangent to the circle at right angles to prove the following theorem:

The angle formed between two tangents meeting at an external point is bisected by a line from the centre of the circle to the external point.

Maa

K

L

O

11 WX is the diameter of a circle with centre at O. Y is a point on the circle and WY is extended to Z so that OY = YZ. Prove that angle ZOX is three times angle YOZ.

O

Y

Z

XW

P QS

R y

O

x

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19.3 Intersecting chords, secants and tangentsIntersecting chordsIn the diagram below, chords PQ and RS intersect at X.

X

Q

S

P

R

• Theorem 6 Code If the two chords intersect inside a circle, then the point of intersection divides each chord into two segments so that the product of the lengths of the segments for both chords is the same.

PX × QX = RX × SX or a × b = c × d

Proof:Join PR and SQ.Consider ΔPRX and ΔSQX.∠PXR = ∠SXQ (vertically opposite angles are equal)∠RSQ = ∠RPQ (angles at the circumference standing on the same arc are equal)∠PRS = ∠PQS 1angles at the circumference standing on the same arc are equal 2ΔPRX ~ ΔSQX (equiangular)

PXSX

= RXQX

(ratio of sides in similar triangles is equal)

or PX × QX = RX × SX

Find the value of the pronumeral.

B

m64 5 D

A

C

X

tHInK WrIte

1 Chords AB and CD intersect at X. Point X divides each chord into two parts so that the products of the lengths of these parts are equal. Write this as a mathematical statement.

AX × BX = CX × DX

2 Identify the lengths of the line segments. AX = 4, BX = m, CX = 6, DX = 5

3 Substitute the given lengths into the formula and solve for m.

4m = 6 × 5

m = 304

= 7.5

WorKeD eXamPLe 4

X

a

c

d

b

Q

S

P

R


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Intersecting secantsIn the diagram below, chords CD and AB are extended to form secants CX and AX respectively. They intersect at X.

B

D

A

C

X

• Theorem 7 Code If two secants intersect outside the circle as shown, then the following relationship is always true:

B

D

A

C

Xdb

a

c

AX × XB = XC × DX or a × b = c × d.

Proof:Join D and A to O, the centre of the circle.Let ∠DCA = x.

∠DOA = 2x (angle at the centre is twice the angle at the circumference standing on the same arc)

Reflex ∠DOA = 360° − 2x (angles in a revolution add to 360°)

∠DBA = 180° − x(angle at the centre is twice the angle at the circumference standing on the same arc)

∠DBX = x(angle sum of a straight line is 180°)∠DCA = ∠DBX

Consider ΔBXD and ΔCXA.∠BXD is common.∠DCA = ∠DBX 1shown previously 2∠XAC = ∠XDB 1angle sum of a triangle is 180° 2ΔAXC ~ ΔDXB 1equiangular 2 AX

DX= XC

XBor AX × XB = XC × DX


B

DA

C

X

6

y

5

7

WorKeD eXamPLe 5

O

C

AB

D

X


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tHInK WrIte

1 Secants XC and AX intersect outside the circle at X. Write the rule connecting the lengths of XC,DX,AX and XB.

XC × DX = AX × XB

2 State the length of the required line segments.

XC = y + 6 DX = 6AX = 7 + 5 XB = 7

= 12

3 Substitute the length of the line segments and solve the equation for y.

(y + 6) × 6 = 12 × 7 6y + 36 = 84 6y = 48 y = 8

Intersecting tangents • In the following diagram, tangents AC and BC intersect at C and AC = BC.

• Theorem 8 Code If two tangents meet outside a circle, then the lengths from the external point to where they meet the circle are equal.Proof:Join A and B to O, the centre of the circle.Consider ΔOCA and ΔOCB.OC is common. OA = OB (radii of the same circle are equal)∠OAC = ∠OBC (radius is perpendicular to tangent through

the point of contact)ΔOCA ≅ ΔOCB (RHS) AC = BC (corresponding sides of congruent

triangles are equal).If two tangents meet outside a circle, the lengths from the external point to the point of contact are equal.


A

m

3

B

C

tHInK WrIte

1 BC and AC are tangents intersecting at C. State the rule that connects the lengths BC and AC.

AC = BC

2 State the lengths of BC and AC. AC = m, BC = 3

3 Substitute the required lengths into the equation to find the value of m.

m = 3

WorKeD eXamPLe 6

B

A

C

B

A

OC


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Chords and radii • In the diagram below, the chord AB and the radius OC intersect at X at 90°; that is,

∠OXB = 90°. OC bisects the chord AB; that is, AX = XB.

• Theorem 9 Code If a radius and a chord intersect at right angles, then the radius bisects the chord.Proof:Join OA and OB.Consider ΔOAX and ΔOBX.

OA = OB (radii of the same circle are equal)∠OXB = ∠OXA (given)OX is common.ΔOAX ≅ ΔOBX (RHS)AX = BX (corresponding sides in congruent triangles are equal)

If a radius and a chord intersect at right angles, then the radius bisects the chord.

• The converse is also true:If a radius bisects a chord, the radius and the chord meet at right angles.

• Theorem 10 Chords equal in length are equidistant from the centre.This theorem states that if the chords MN and PR are of equal length, then OD = OC.Proof:Construct OA ⟂ MN and OB ⟂ PR.Then OA bisects MN and OB bisects PR (Theorem 9)Because MN = PR, MD = DN = PC = CR.Construct OM and OP, and consider ΔODM and ΔOCP. MD = PC (shown above) OM = OP (radii of the same circle are equal)∠ODM = ∠OCP = 90° (by construction)ΔODM ≅ ΔOCP (RHS) So OD = OC (corresponding sides in congruent triangles are equal)Chords equal in length are equidistant from the centre.

Find the values of the pronumerals, given that AB = CD.

B

C

A

O

mn

G

F

E

D2.5

3

H

WorKeD eXamPLe 7

BXA

C

O

BXA

C

O

B

N

AC

O

M P

RD

B

N

AC

O

M P

RD


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tHInK WrIte

1 Since the radius OG is perpendicular to the chord AB, the radius bisects the chord.

AE = EB

2 State the lengths of AE and EB. AE = m, EB = 33 Substitute the lengths into the equation to fi nd the

value of m.m = 3

4 AB and CD are chords of equal length and OE and OF are perpendicular to these chords. This implies that OE and OF are equal in length.

OE = OF

5 State the lengths of OE and OF. OE = n, OF = 2.5

6 Substitute the lengths into the equation to fi nd the value of n.

n = 2.5

The circumcentre of a triangle • In the diagram, a circle passes through the vertices of the triangle ABC.

C

BA

• The circle is called the circumcircle of triangle ABC, and the centre of the circle is called the circumcentre.

• The circumcentre is located as follows.Draw any triangle ABC. Label the vertices.

C

BA

Construct perpendicular bisectors of AB,AC and BC, and let the bisectors intersect at O. This means that OA = OB = OC, so a circle can be drawn through A,B and C with a centre at O.

C

O

BA

Exercise 19.3 Intersecting chords, secants and tangents InDIVIDuaL PatHWays

⬛PraCtIseQuestions:1–7, 10

⬛ConsoLIDateQuestions:1–8, 10


reFLeCtIon What techniques will you use to prove circle theorems?


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FLuenCy

1 WE4 Find the value of the pronumeral in each of the following.a A D

BC

X

m 4

86

b A

D

B

C

Xm

629

c

A

D B

C

X

m

m

49

2 WE5 Find the value of the pronumeral in each of the following.a

n

42

3

b

m 4.5

5 6

c

n

48

3

d

65

m7

3 WE6 Find the value of the pronumerals in each of the following.a

x

5b

m

7

c

y

x3.1

2.5

4 WE7 Find the value of the pronumeral in each of the following.a

Ox

2.8

b 3.3

O

xc

2.52.55.6O

x

d

mO

unDerstanDIng

5 MC Note: There may be more than one correct answer.In which of the following figures is it possible to find the value of m through solving a linear equation?a

2

5

7

m

B

2

3

7

m

C

2

3

4

m

D

21

4 m


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6 Find the length, ST, in the diagram below.

5 cm4 cm

9 cmP

Q

ST

R

reasonIng

7 Prove the result: If a radius bisects a chord, then the radius meets the chord at right angles. Remember to provide reasons for your statements.

8 Prove the result: Chords that are an equal distance from the centre are equal in length. Provide reasons for your statements.

9 Prove that the line joining the centres of two intersecting circles bisects their common chord at right angles. Provide reasons for your statements.

ProBLem soLVIng

10 Calculate the pronumeral for each of the following diagrams.a

x

6

3b

x

8

4

10

c

y4x

3x

15

O

11 AOB is the diameter of the circle. CD is a chord perpendicular to AB and meeting AB at M.

b

c

M

OA B

C

D

a

a Why is M the midpoint of CD?b If CM = c, AM = a and MB = b, prove that c2 = ab.

c Explain why the radius of the circle is equal to a + b

2.


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CHaLLenge 19.1CHaLLenge 19.1CHaLLenge 19.1CHaLLenge 19.1CHaLLenge 19.1CHaLLenge 19.1

19.4 Cyclic quadrilateralsQuadrilaterals in circles • A cyclic quadrilateral has all four vertices on the circumference of

a circle; that is, the quadrilateral is inscribed in the circle.In the diagram at right, points A, B, C and D lie on the circumference;

hence, ABCD is a cyclic quadrilateral.It can also be said that points A, B, C and D are concyclic; that is, the circle passes

through all the points.

• Theorem 11 Code The opposite angles of a cyclic quadrilateral are supplementary (add to 180°).Proof:Join A and C to O, the centre of the circle.Let ∠ABC = x.Refl ex ∠AOC = 2x ( angle at the centre is twice the angle at the circumference standing

on the same arc)Refl ex ∠AOC = 360° − 2x (angles in a revolution add to 360°)

∠ADC = 180° − x ( angle at the centre is twice the angle at the circumference standing on the same arc)

∠ABC + ∠ADC = 180°Similarly, ∠DAB + ∠DCB = 180°.Opposite angles in a cyclic quadrilateral are supplementary.

• The converse is also true:If opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic.

Find the values of the pronumerals in the diagram below. Give reasons for your answers.

P

R

Q

S

75°120°

x

y

WorKeD eXamPLe 8

A

C

B

D

A

C

B

D

O


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tHInK WrIte

1 PQRS is a cyclic quadrilateral, so its opposite angles are supplementary. First fi nd the value of x by considering a pair of opposite angles ∠PQR and ∠RSP and forming an equation to solve.

∠PQR + ∠RSP = 180° (The opposite angles of a cyclic quadrilateral are supplementary.) ∠PQR = 75°,∠RSP = xx + 75° = 180° x = 105°

2 Find the value of y by considering the other pair of opposite angles (∠SPQ and ∠QRS).

∠SPQ + ∠QRS = 180° ∠SPQ = 120°,∠QRS = y y + 120° = 180° y = 60°

• Theorem 12 Code The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.Proof:∠QPS + ∠QRS = 180° (opposite angles of a cyclic quadrilateral)∠QPS + ∠SPT = 180° (adjacent angles on a straight line)Therefore ∠SPT = ∠QRS.The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

Find the value of the pronumerals in the diagram below.

D B

A

C

y

x100°

50°

tHInK WrIte

1 ABCD is a cyclic quadrilateral. The exterior angle, x, is equal to its interior opposite angle, ∠DAB.

x = ∠DAB, ∠DAB = 50° So x = 50°.

2 The exterior angle, 100°, is equal to its interior opposite angle, ∠ADC.

∠ADC = 100°, ∠ADC = y So y = 100°.

WorKeD eXamPLe 9

Exercise 19.4 Cyclic quadrilaterals InDIVIDuaL PatHWays

⬛PraCtIseQuestions:1–6, 8

⬛ConsoLIDateQuestions:1–8


P

R

Q

S

b

aaT

reFLeCtIon What is a cyclic quadrilateral?


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FLuenCy

1 WE8 Find the values of the pronumerals in each of the following.a

yx

92°65°

b

m95°

c n

155°

d x

50°

e

x yO

f

x

y

O135°

85°

2 WE9 Find the values of the pronumerals in each of the following.a

x

y80°

85°

b x

y

115°

110°

c

x

95°

d x

150°

e x

y 120°

f

mn

120° 130°

3 MC Note: There may be more than one correct answer.Which of the following correctly states the relationship between x,y and z in the diagram shown?a x = y and x = 2zC z = 2x and y = 2z

B x = 2y and y + z = 180°D x + y = 180° and z = 2x

unDerstanDIng

4 The steps below show you how to set out the proof that the opposite angles of a cyclic quadrilateral are equal.a Find the size of ∠DOB.b Find the size of the refl ex angle DOB.c Find the size of ∠BCD.d Find ∠DAB + ∠BCD.

5 MC Note: There may be more than one correct answer.a Which of the following statements is always true for the diagram

shown?a r = t B r = p C r = q D r = s

b Which of the following statements is correct for the diagram shown?a r + p = 180° B q + s = 180°C t + p = 180° D t = r

doc-5396doc-5396doc-5396

z

x

y

O

xA

D

B

C

O

s

q r

t p


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reasonIng

6 Prove that the exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

7 Calculate the values of the pronumerals in each of these diagrams.a

yx

z 87°

b

16x2 – 10x

20x2 – 8x

c

2x – 1 – 3x2

3x2 – 5x + 1

ProBLem soLVIng

8 Calculate the value of each pronumeral in the diagram at right.

9 ∠FAB = 70°, ∠BEF = a°, ∠BED = b° and ∠BCD = c°.

A

F

E

B

C

D

70°

a°

c°

b°

a Find the values of a, b and c.b Prove that CD is parallel to AF.

110°

110°

x

w

y2z

z+ 5

doc-14628doc-14628doc-14628


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19.5 Tangents, secants and chordsThe alternate segment theorem • Consider the figure shown. Line BC is a tangent to the circle

at the point A. • A line is drawn from A to anywhere on the circumference,

point D.The angle ∠BAD defines a segment (the shaded area).The unshaded part of the circle is called the alternate segment to ∠BAD.Now consider angles subtended by the chord AD in the alternate segment, such as the angles marked in red and blue.

• The alternate segment theorem states that these are equal to the angle that made the segment, namely:

∠BAD = ∠AED and ∠BAD = ∠AFD

• Theorem 13 Code The angle between a tangent and a chord is equal to the angle in the alternate segment.

Proof:We are required to prove that ∠BAD = ∠AFD.Construct the diameter from A through O, meeting the circle at G.Join G to the points D and F.

∠BAG = ∠CAG = 90°(radii ⟂ tangent at point of contact)∠GFA = 90° (angle in a semicircle is 90°)∠GDA = 90° (angle in a semicircle is 90°)

Consider ΔGDA. We know that ∠GDA = 90°.∠GDA + ∠DAG + ∠AGD = 180°

90° + ∠DAG + ∠AGD = 180°∠DAG + ∠AGD = 90°

∠BAG is also a right angle.

∠BAG = ∠BAD + ∠DAG = 90°

Equate the two results.

∠DAG + ∠AGD = ∠BAD + ∠DAG

Cancel the equal angles on both sides.

∠AGD = ∠BAD

Now consider the fact that both triangles DAG and DAF are subtended from the same chord (DA).

∠AGD = ∠AFD (Angles in the same segment standing on the same arc are equal).Equate the two equations.

∠AFD = ∠BAD

OD

A CB

OD

E

F

A CB

OD

G

F

A CB


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Find the value of x and y, giving reasons.

x

y62°D

B

A

C

T

tHInK WrIte

1 Use the alternate segment theorem to find x. x = 62° (angle between a tangent and a chord is equal to the angle in the alternate segment)

2 The value of y is the same as x because x and y are subtended by the same chord BT.

y = 62° (angles in the same segment standing on the same arc are equal)

WorKeD eXamPLe 10

Tangents and secants • Theorem 14 Code

If a tangent and a secant intersect as shown, the following relationship is always true:XA × XB = (XT)2 or a × b = c2.

Proof:Join BT and AT.Consider ΔTXB and ΔAXT.

∠TXB is common.∠XTB = ∠XAT (angle between a tangent and a chord is equal

to the angle in the alternate segment)∠XBT = ∠XTA (angle sum of a triangle is 180°)ΔTXB~ΔAXT (equiangular)

So XBXT

= XTXA

or XA × XB = (XT)2.


8T

B

A

X

m5

tHInK WrIte

1 Secant XA and tangent XT intersect at X. Write the rule connecting the lengths of XA, XB and XT.

XA × XB = (XT)2

WorKeD eXamPLe 11

c T

B

A

Xb

a

T

B

A

X


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2 State the values of XA, XB and XT. XA = m + 5, XB = 5, XT = 8

3 Substitute the values of XA, XB and XT into the equation and solve for m.

(m + 5) × 5 = 82

5m + 25 = 64 5m = 39

m = 7.8

Exercise 19.5 Tangents, secants and chords InDIVIDuaL PatHWays

⬛PraCtIseQuestions:1, 2, 4–6, 8, 10, 13–15, 17, 20, 22

⬛ConsoLIDateQuestions:1–3, 5, 7, 9, 11, 13, 14, 16, 19, 21, 22


FLuenCy

1 WE10 Find the value of the pronumerals in the following.a

x

70°

b

xy

59°

47°

2 WE11 Find the value of the pronumerals in the following.a

p

54

b

q

12

4

3 Line AB is a tangent to the circle as shown in the fi gure on the right. Find the values of the angles labelled x and y.

Questions 4 to 6 refer to the fi gure on the right. The line MN is a tangent to the circle, and EA is a straight line. The circles have the same radius.

4 Find 6 different right angles.5 MC If ∠DAC = 20°, then ∠CFD and ∠FDG are respectively:

a 70° and 50° B 70° and 40° C 40° and 70° D 70° and 70°6 MC A triangle similar to FDA is:

a FDG B FGB C EDA D GDE

reFLeCtIon Describe the alternate segment of a circle.

B

A

y

x

21°

O

D M

B N

FE AGO C


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7 Find the values of the angles x and y in the figure at right.

unDerstanDIng

8 Show that if the sum of the two given angles in question 7 is 90°, then the line AB must be a diameter.

9 Find the value of x in the figure at right, given that the line underneath the circle is a tangent.

10 In the figure at right, express x in terms of a and b. This is the same drawing as in question 9.

11 Two tangent lines to a circle meet at an angle y, as shown in the figure at right. Find the values of the angles x,y and z.

12 Solve question 11 in the general case (see the figure at right) and show that y = 2a. This result is important for space navigation (imagine the circle to be the Earth) in that an object at y can be seen by people at x and z at the same time.

13 In the figure at right, find the values of the angles x,y and z.

14 MC Examine the figure at right. The angles x and y (in degrees) are respectively:a 51 and 99B 51 and 129C 39 and 122D 51 and 122

A

B

42°62°

y

xO

100°

20°

x

O

x

Oa

b

x

z

10°

y

O

x

z

a

y

O

20°

75°z

xy

O

51

19y

x O


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Questions 15 to 17 refer to the figure at right. The line BA is a tangent to the circle at point B. Line AC is a chord that meets the tangent at A.

15 Find the values of the angles x and y.

16 MC The triangle which is similar to triangle BAD is:a CAB B BCD C BDC D AOB

17 MC The value of the angle z is:a 50° B 85° C 95° D 100°

reasonIng

18 Find the values of the angles x,y and z in the figure at right. The line AB is tangent to the circle at B.

19 Find the values of the angles x,y and z in the figure at right. The line AB is tangent to the circle at B. The line CD is a diameter.

20 Solve question 19 in the general case; that is, express angles x,y and z in terms of a (see the figure at right).

21 Prove that, when two circles touch, their centres and the point of contact are collinear.

ProBLem soLVIng

22 Find the value of the pronumerals in the following.a

4

6

x

b

4

8

k

c

4

5

mn

d 7

x

C

D

BA50° 45°z

xy

O

xzy

A

DC

B

92°33°

O

x

yz

C

A

D

B25°

O

x

y

z

C

A

D

Ba

O


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e 1

5.5

811

6 b

a f

3

2

w

x

23 Find the values of a, b and c in each case.a ∠BCE = 50° and ∠ACE = c b E

A

D50°

70° bac

F

CB

CHaLLenge 19.2CHaLLenge 19.2CHaLLenge 19.2CHaLLenge 19.2CHaLLenge 19.2CHaLLenge 19.2

C

E

D

A

B

acb

50°


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Link to assessON for questions to test your readiness For learning, your progress as you learn and your levels oF achievement.

assessON provides sets of questions for every topic in your course, as well as giving instant feedback and worked solutions to help improve your mathematical skills.

www.assesson.com.au

int-2880int-2880int-2880

int-2881int-2881int-2881

int-3894int-3894int-3894

LanguageLanguageLanguage

alternate segment theoremalternate segment theoremalternate segment theoremangleangleanglearcarcarcchordchordchordchordchordchordcirclecirclecirclecircumcentrecircumcentrecircumcentrecircumcirclecircumcirclecircumcircle

circumferencecircumferencecircumferenceconcyclicconcyclicconcycliccycliccycliccycliccyclic quadrilateralcyclic quadrilateralcyclic quadrilateraldiameterdiameterdiametermajor segmentmajor segmentmajor segmentminor segmentminor segmentminor segment

radiusradiusradiussecantsecantsecantsectorsectorsectorsegmentsegmentsegmentsegmentsegmentsegmentsubtendsubtendsubtendtangenttangenttangenttheoremtheoremtheorem

ONLINE ONLY 19.6 ReviewThe Maths Quest Review is available in a customisable format for students to demonstrate their knowledge of this topic.

The Review contains:• Fluency questions — allowing students to demonstrate the

skills they have developed to effi ciently answer questions using the most appropriate methods

• Problem solving questions — allowing students to demonstrate their ability to make smart choices, to model and investigate problems, and to communicate solutions effectively.

A summary of the key points covered and a concept map summary of this topic are available as digital documents.

Review questionsDownload the Review questions document from the links found in your eBookPLUS.

www.jacplus.com.au

the story of mathematicsis an exclusive Jacaranda video series that explores the history of mathematics and how it helped shape the world we live in today.

Piscopia (eles-2022) tells the story of Elena Cornaro Piscopia, a female mathematician who challenged the strict ways of the Catholic Church in 17th Century Italy to become the fi rst woman to be awarded a university doctorate degree.


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836 Maths Quest 10 + 10A836 Maths Quest 10 + 10A

<INVESTIGATION> FOR RICH TASK OR <MEASUREMENT AND GEOMETRY> FOR PUZZLE

Variation of distance

Equator

15°N

30°N

45°N

60°N

15°S

30°S

45°S

60°S

135°

E

120°

E

105°

E90°E

75°E

150°

E

165°

E

180°

165°

W

135°

E 30°N

45°N

Latitude

Longitude

Equator

15°N

30°N

45°N 60°N

15°S

30°S

45°S60°S

135°

E15

0°E

165°

E18

0°16

5°W

120°

E

105°

E

90°E75

°E

North Pole

South Pole

Equator

P1 P2

100E0

0

RICH TASK

INVESTIGATION

The distance (in km) between two points on the same line of latitude is given by the formula:

Distance = angle sector between the two points × 111 × cos 1degree of latitude 2 .


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MEASUREMENT AND GEOMETRY


1 The size of the angle sector between P1 and P2 is 100° and these two points lie on 0° latitude. The distance between the points would be calculated as 100 × 111 × cos 0°. Determine this distance.

2 Move the two points to the 10° line of latitude. Calculate the distance between P1 and P2 in this position. Round your answer to the nearest kilometre.

3 Complete the following table showing the distance (rounded to the nearest kilometre) between the points P1 and P2 as they move from the equator towards the pole.

4 Describe what happens to the distance between P1 and P2 as we move from the equator to the pole. Is there a constant change? Explain.

5 You would perhaps assume that, at a latitude of 45°, the distance between P1 and P2 is half the distance between the points at the equator. This is not the case. At what latitude does this occur?

6 On the grid lines provided, sketch a graph displaying the change in distance between the points in moving from the equator to the pole.

7 Consider the points P1 and P2 on lines of longitude separated by 1°. On what line of latitude (to the nearest degree) would the points be 100 km apart?

8 Keeping the points P1 and P2 on the same line of latitude, and varying their lines of longitude, investigate the rate that the distance between them changes from the equator to the pole. Is it more or less rapid in comparison to what you found earlier?


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838 Maths Quest 10 + 10A838 Maths Quest 10 + 10A

<InVestIgatIon> For rICH tasK or <measurement anD geometry> For PuZZLe

Where is 2 litres of hydrochloric acid produced each day?The values of the lettered angles give the puzzle’s answer code.

115º88º 100º80º125º111º 90º88º73º95º 110º50º105º

120º105º 88º125º105º110º 105º117º99º88º 115º99º115º 95º

80º85º

75º

I

D

R107º

81º

H

100º

115º

125º

S

C

T117º60º

E

G

69º

92º

YO

110º

A

130º

U MN

CoDe PuZZLe



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Activities19.1 overviewVideo• The story of mathematics (eles-2022)

19.2 angles in a circleInteractivity• Angles in a circle (int-2795)Digital docs• SkillSHEET (doc-5390): Using tests to prove congruent

triangles • SkillSHEET (doc-5391): Corresponding sides and

angles of congruent triangles • SkillSHEET (doc-5392): Using tests to prove similar

triangles • SkillSHEET (doc-5393): Angles in a triangle • SkillSHEET (doc-5394): More angle relations • WorkSHEET 19.1 (doc-14627): Circle geometry I

19.4 Cyclic quadrilateralsDigital docs• SkillSHEET (doc-5396): Angles in a quadrilateral• WorkSHEET 19.2 (doc-14628): Circle geometry II

19.6 reviewInteractivities• Word search (int-2880)• Crossword (int-2881)• Sudoku (int-3894)Digital docs• Topic summary (doc-14629)• Concept map (doc-14630)

to access eBookPLus activities, log on to www.jacplus.com.au



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Exercise 19.2 — Angles in a circle 1 a x = 30° (theorem 2)

b x = 25°, y = 25° (theorem 2 for both angles)c x = 32° (theorem 2)d x = 40°, y = 40° (theorem 2 for both angles)e x = 60° (theorem 1)f x = 40° (theorem 1)g x = 84° (theorem 1)h x = 50° (theorem 2); y = 100° (theorem 1)i x = 56° (theorem 1)

2 a s = 90°, r = 90° (theorem 3 for both angles)b u = 90° (theorem 4); t = 90° (theorem 3)c m = 90°, n = 90° (theorem 3 for both angles)d x = 52° (theorem 3 and angle sum in a triangle = 180°)e x = 90° (theorem 4)f x = 90°(theorem 4); y = 15° (angle sum in a triangle = 180°)

3 a x = z = 90° (theorem 4); y = w = 20° (theorem 5 and angle sum in a triangle = 180°)

b s = r = 90° (theorem 4); t = 140° (angle sum in a quadrilateral = 360°)

c x = 20° (theorem 5); y = z = 70° (theorem 4 and angle sum in a triangle = 180°)

d s = y = 90° (theorem 4); x = 70° (theorem 5); r = z = 20° (angle sum in a triangle = 180°)

e x = 70° (theorem 4 and angle sum in a triangle = 180°); y = z = 20° (angle sum in a triangle = 180°)

f x = y = 75° (theorem 4 and angle sum in a triangle = 180°); z = 75° (theorem 1)

4 D5 B, D6 a Base angles of a right-angled isosceles

triangleb r + s = 90°, s = 45°⇒ r = 45°c u is the third angle in ΔABD, which is right-angled.d m is the third angle in ΔOCD, which is right-angled.e ∠AOC and ∠AFC stand on the same arc with ∠AOC at the

centre and ∠AFC at the circumference.7 OR = OP (radii of the circle)

∠OPR = x (equal angles lie opposite equal sides)∠SOP = 2x (exterior angle equals the sum of the two interior opposite angles)OR = OQ (radii of the circle)∠OQR = y (equal angles lie opposite equal sides)∠SOQ = 2y (exterior angle equals the sum of the two interior opposite angles)Now ∠PRQ = x + y and ∠POQ = 2x + 2y = 2(x + y).Therefore ∠POQ = 2 × ∠PRQ.

8, 9 Check with your teacher.10 Check with your teacher.11 Check with your teacher.

Exercise 19.3 — Intersecting chords, secants and tangents 1 a m = 3 b m = 3 c m = 6 2 a n = 1 b m = 7.6 c n = 13 d m = 4 3 a x = 5 b m = 7 c x = 2.5,y = 3.14 a x = 2.8 b x = 3.3 c x = 5.6 d m = 90°5 B, C, D6 ST = 3cm

7, 8, 9 Check with your teacher.10 a x = 3"2 b x = 6 c x = 3, y = 1211 Check with your teacher.

Challenge 19.1

Exercise 19.4 — Cyclic quadrilaterals 1 a x = 115°,y = 88° b m = 85°

c n = 25° d x = 130°e x = y = 90° f x = 45°, y = 95°

2 a x = 85°, y = 80° b x = 110°, y = 115°c x = 85° d x = 150°e x = 90°, y = 120° f m = 120°, n = 130°

3 D4 a 2x b 360° − 2x

c 180° − x d 180°5 a A

b A, B, C, D6 Check with your teacher.7 a x = 93°, y = 87°, z = 93°

b x = −2° or 52°

c x = 23° or 1

2°

8 w = 110°, x = 70°, y = 140°, z = 87.5°9 a a = 110°, b = 70° and c = 110°

b Check with your teacher.

Exercise 19.5 — Tangents, secants and chords 1 a x = 70°

b x = 47°, y = 59° 2 a p = 6

b q = 83 x = 42°, y = 132°4 MAC,NAC,FDA,FBA,EDG,EBG5 B6 D7 x = 42°, y = 62°8 Answers will vary.9 60°

10 x = 180° − a − b11 x = 80°,y = 20°,z = 80°12 Answers will vary.13 x = 85°, y = 20°, z = 85°14 D15 x = 50°, y = 95°16 A17 C18 x = 33°, y = 55°, z = 22°19 x = 25°, y = 65°, z = 40°20 x = a, y = 90° − a, z = 90° − 2a21 Check with your teacher. 2 2 a x = 5 b k = 12 c m = 6, n = 6

d x = 7 e b = 4, a = 2 f w = 3, x = 523 a a = 50°, b = 50° and c = 80°

b a = 50°, b = 70° and c = 70°

AnswerstoPIC 19 Circle geometry


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Challenge 19.2

3

1 2 4

5

6

7

Investigation — Rich task1 11 100 km2 10 931 km3

Latitude Distance between P1 and P2 (km)

0° 11 100

10° 10 931

20° 10 431

30° 9613

40° 8503

50° 7135

60° 5550

70° 3796

80° 1927

90° 0

4 The distance between P1 and P2 decreases from 11 100 km at the equator to 0 km at the pole. The change is not constant. The distance between the points decreases more rapidly on moving towards the pole.

5 Latitude 60°6 12 000

11 00010 000

900080007000600050004000300020001000

0

Dis

tanc

e be

twee

n P 1 a

nd P

2 (km

)

10° 20° 30° 40°Latitude

50° 60° 70° 80° 90°

7 Latitude 26°8 Answers will vary. Teacher to check.

Code puzzleIn your stomach to aid digestion


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Documents

Circle geometry - Wiley: · PDF file812 Maths Quest 10 + 10A measurement anD geometry The angle subtended at the centre of a circle is twice the angle subtended at the circumference,