measurement and geometrycircle geometry...

15Measurement and geometry

CirclegeometryOptional Stage 5.3 topicThe ancient Greek philosopher, Thales of Miletus(624�546 BCE), is believed to have been the firstmathematician and scientist to attempt to explain thingsby reasoning and deduction rather than by resorting tomythology and superstition. He lived before Pythagorasand proved the first theorem in mathematics: that anyangle drawn on the circumference of a semicircle is a rightangle. This rule is often called Thales’ theorem.

n Chapter outlineProficiency strands

15-01 Parts of a circle* U C15-02 Chord properties of

circles* U F R C15-03 Angle properties of

circles* U F R C15-04 Tangent and secant

properties of circles* U F R C15-05 Proofs using circle

theorems* U F PS R C

*STAGE 5.3

nWordbankcollinear points Points that lie on a straight line.

converse A rule or statement turned back-to-front; thereverse statement

cyclic quadrilateral A quadrilateral bounded by a circle,whose vertices lie on the circle’s circumference

intercept The distance between the points where a lineintersects other lines or curves

perpendicular bisector An interval or line that bisectsanother interval at right angles

secant A line that intersects a curve at two points

subtend an angle To sit opposite an angle and ‘hold’ itsarms

tangent A line that touches a curve at one point butdoes not cross it

Shut

ters

tock

.com

/Ron

Elli

s

9780170194662

NEW CENTURY MATHS ADVANCEDfor the A u s t r a l i a n C u r r i c u l u m10þ10A

n In this chapter you will:• (STAGE 5.3) Prove and apply angle and chord properties of circles• (STAGE 5.3) identify the parts of a circle• (STAGE 5.3) prove chord and angle properties of circles and use them to find unknown

lengths and angles in circles• (STAGE 5.3) prove tangent and secant properties of circles and use them to find unknown

lengths and angles in circles

SkillCheck

1 Which congruence test (SSS, SAS, AAS or RHS) can be used to prove that each pair oftriangles are congruent?

cba

fed

6 cm

10 cm

10 cm

6 cm

2 Which similarity test (‘SSS’, ‘SAS’, ‘AA’ or ‘RHS’) can be used to prove that each pair oftriangles are similar?

cba

fed

1420

21

15

30

10 9

12 20

15

3

3.75

4.5

3

2

2.5 43

6

8

110°

30°40°

95° 30°

30°

55°

Worksheet

StartUp assignment 15

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Chapter 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Circle geometry

15-01 Parts of a circle

The radius is an interval joining the centre to anypoint on the circle.The circumference is the perimeter of a circle.

radius

centre

circumference

A tangent is a line touching the circle atone point.A secant is a line that cuts the circle attwo points.

secant

tangent

A semicircle is half of a circle.

semicircle

A quadrant is quarter of a circle.

Quadrant

A chord is an interval joining two points on the circle.A diameter is a chord that passes through thecentre.An arc is a part of the circumference of the circle.

chord

diameter

arc

A sector is part of a circle bounded bytwo radii and an arc.A segment is part of a circle cut off bya chord.

segment

sector

Exercise 15-01 Parts of a circle1 Name each part of the circle marked by a letter.

a

b

d

c

Stage 5.3

Worksheet

Parts of a circle

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5599780170194662


2 a Draw a sector and a segment in a circle.

b What is the difference between a sector and a segment?

3 State the mathematical rule connecting d (diameter) and r (radius) as an algebraic formula.

4 Which parts of a circle are labelled a and b? Select the correct answerA, B, C or D. a

bA diameter and segment B radius and segmentC diameter and sector D radius and sector

5 Write the correct word to match each description.a the distance from the centre of a circle to its side

b quarter of a circle

c line that touches the outside of a circle once

d an interval from one side of a circle to the other side, through the circle’s centre

e an interval from one side of the circle to the other side, not through the centre

f part of the circumference of a circle

g the area inside a circle formed by two radii and an arc

h the complete distance around a circle

i the area inside a circle formed by a chord and an arc

6 Name each part of the circle marked by a letter.

a

c

b d

Technology Perpendicular to a chord1 Click View and Axes in GeoGebra.

2 Draw a circle using Circle with centre through point.3 Insert one new point on the circumference of the circle.

4 Use Interval between two points to join BC.

5 Click Midpoint or Centre and chord BC.

6 Select Perpendicular line and A and the midpoint of chord BC.

7 Select Distance or Length and measure C to the midpoint of BC and repeat for point B tothe midpoint of BC. Use the Move tool to manipulate points B and C. What do you notice?

Stage 5.3

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Chapter 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Circle geometry

15-02 Chord properties of circles

Equal chordsWhen an arc or chord of a circle subtends an angle, it sits opposite or under the angle and ‘holds’the angle’s arms. The angle stands on the ends of the arc or chord. The angle is subtended by thearc or chord.

A

O

B

P

X

Y

The arc AB subtends the angle at O. The chord XY subtends the angle at P.

A theorem is a rule or statement that can be proved from basic principles. We will now look attheorems relating to chords that can be proved by congruent triangles. These theorems can also bedemonstrated using GeoGebra, other graphing technology, or pencil-and-paper constructions.

Summary

Theorem 1: Chords of equal length in a circle subtend equal angles atthe centre of the circle.

O

AB

D

C

If AB = CD,∠AOB = ∠COD.

Proof:In 4AOB and 4COD:AB ¼ CD (given)OA ¼ OC (equal radii)OB ¼ OD (equal radii)[ 4AOB ” 4COD (SSS)[ \AOB ¼ \COD (matching angles of congruent triangles)So chords of equal length subtend equal angles at the centre of a circle.

Summary

Theorem 2: Chords of equal length in a circle are equidistant from thecentre of the circle.

C

D

X

Y

A B

O

If AB = CD,then OX = OY.

Stage 5.3

Worksheet

A page of circles

MAT10MGWK10234

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Proof:Draw radii OA, OB, OC and OD.

BX

Y

A

CD

OIn 4AOX and 4COY:OA ¼ OC (equal radii)\A ¼ \C (matching angles of congruent 4AOB and 4COD proved in

Theorem 1)\OXA ¼ \OYC ¼ 90� (given)[ 4AOX ” 4COY (AAS)[ OX ¼ OY (matching sides of congruent triangles)So equal chords are equidistant from the centre of a circle.

Perpendicular bisectors of chords

Summary

Theorem 3: The perpendicular from the centre of a circle to achord bisects the chord.

O

A X B

The converse of this is also true. The line from the centre ofa circle to the midpoint of a chord is perpendicular to thechord.

The word ‘converse’ means to ‘turn around’ or ‘reverse’. So the converse of a theorem is thetheorem written ‘back-to-front’.Proof:Draw radii OA and OB.

BX

A

OIn 4AOX and 4BOX:OA ¼ OB (equal radii)OX is common.\OXA ¼ \OXB ¼ 90� (OX ’ AB)[ 4AOX ” 4BOX (RHS)[ AX ¼ BX (matching sides of congruent triangles)So the perpendicular from the centre of a circle to a chord bisects the chord.

Summary

Theorem 4: The perpendicular bisector of a chord in a circle passesthrough the centre of the circle.

BX

AO

Stage 5.3

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Chapter 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Circle geometry

Proof:The perpendicular from O to AB bisects AB. (proved in Theorem 3)[ OX is the perpendicular bisector of AB.[ The perpendicular bisector of AB passes through the centre O.

Common chords

Summary

Theorem 5: When two circles intersect, the line joining theircentres bisects their common chord at right angles.

O

A

B

P

Proof:Draw radii OA, OB, PA and PB and let OP and AB bisect at M.

O M

A

B

PIn 4OAP and 4OBP:OA ¼ OB (equal radii)PA ¼ PB (equal radii)OP is common.[ 4OAP ” 4OBP (SSS)[ \AOM ¼ \BOM (matching angles of congruent triangles)In 4OAM and 4OBM:OA ¼ OB (equal radii)OM is common.[ \AOM ¼ \BOM (proved above)[ 4OAM ” 4OBM (SAS)[ AM ¼ MB (matching sides of congruent triangles)[ AB ’ OP (line from the centre of a circle to the midpoint of a chord)So the perpendicular from the centre of a circle to a chord bisects the chord.

Example 1

In each diagram, O is the centre of a circle.

a If PQ ¼ 35 cm, find PT. b If OP ¼16 cm, find AB.

TP

QO

P

B

A O1634

Stage 5.3

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Solution

a PT ¼ 12ðPQÞ ðthe perpendicular from the centre bisects the chordÞ

¼ 12

3 35

¼ 17:5 cm

b PB2 ¼ 342 � 162

¼ 900

PB ¼ffiffiffiffiffiffiffiffi

900p

¼ 30 cm

by Pythagoras’ theorem

) AB ¼ 2 3 PB ðthe perpendicular from the centre bisects the chordÞ¼ 60 cm

Example 2

The length AB of the common chord of two circles is 24 cm.If the radii of the circles are 13 cm and 15 cm, find thedistance between their centres.

A

B

X Y13 cm 15 cm

AB = 24 cm

SolutionWe need to find XY.

Let XY and AB cross at M.

XY ’ AB (common chord of circles)

AM ¼ 12

3 AB

¼ 12

3 24

¼ 12 cm

A

B

MX Y

XM2 ¼ 132 � 122

¼ 25

XM ¼ffiffiffiffiffi

25p

¼ 5 cm

by Pythagoras’ theorem

YM 2 ¼ 152 � 122

¼ 81

YM ¼ffiffiffiffiffi

81p

¼ 9 cm

) XY ¼ 5þ 9

¼ 14 cmThe distance between the centres is 14 cm.

Stage 5.3

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Chapter 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Circle geometry

Exercise 15-02 Chord properties of circles1 The converse of Theorem 1 is: ‘Chords subtending equal angles at the

centre of a circle have equal lengths’.

A

C

O

B

D

∠ AOB = ∠ COD

a Prove that 4AOB ” 4COD, given that \AOB ¼ \COD.

b Hence, prove that AB ¼ CD.

2 The converse of Theorem 2 is: ‘Chords that are equidistant from thecentre of a circle are equal’.

X

Y

C

A

O

B

D

OX = OY

a Prove that 4OXA ” 4OYC and therefore AX ¼ CY.

b Prove that 4OXB ” 4OYD and therefore BX ¼ DY.

c Hence show that AB ¼ CD.

3 The converse of Theorem 3 is: ‘The line from the centre of a circle tothe midpoint of a chord is perpendicular to the chord.’

O

ABX

AX = BX

a Prove that 4OXA ” 4OXB and therefore \OXA ¼ \OXB.

b Prove that OX ’ AB.

4 a Draw a circle and construct any two non-parallel chords.

b Bisect these chords.

c Through what point must each perpendicular bisector pass? (Use Theorem 4.)

d Explain why the point of intersection between the two perpendicular bisectors is the centreof the circle.

5 The diagram shows the perpendicular bisectors of two sides AB andCA of a triangle, intersecting at O. It can be proved that O is thecentre of a circle passing through all three vertices of the triangle.

A

C

O

B

Y

X

The full statement on this theorem is: ‘Given any three non-collinear* points, the point ofintersection of the perpendicular bisectors of any two sides of the triangle formed by the threepoints is the centre of the circle through all three points.(*Non-collinear means not on the same straight line.)Prove this theorem as follows:a prove that 4AOY ” 4COY and therefore OA ¼ OC

b prove that 4AOX ” 4BOX and therefore OA ¼ OB

c hence prove that the circle with centre O and radius OA also passes through B and C.

6 This circle has a radius of 7 cm and the chord RS is 8 cm. Find,correct to 2 decimal places, the length of OA.

R

A

S

O

Stage 5.3

See Example 1

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7 In the diagram, AB ’ OP. If OB ¼ 32.5 cm and OP ¼ 12.5 cm,find AB. A

P

B

O

8 In each problem, state which chord theorem you use.

a OU ¼ OV

AB ¼ 9 mFind the length of UC.

b GF ¼ 12 mFind the length of DE.

c UV ¼ VW

\VOW ¼ 64�Find the size of \UVO.

D BO

U V

A C

OD

E

G

F

V

O

U W

d OT ’ PQ

OQ ¼ 17 mmOT ¼ 8 mmFind PQ.

e OM ’ XY

XY ¼ 40 cmOX ¼ 29 cmFind OM.

f DE ’ ON

ON ¼ 18 mDE ¼ 36 mFind the radius.

O

P QT

O M

X

Y

O

D EN

9 Find the radius of a circle in which a chord of length 96 cm is 20 cm from the centre.

10 On a clear day, you can see 11.5 km in any direction from a particular lookout. If a straightroad is 6.9 km from the lookout, what length of the road can be seen from the lookout?

11 In the diagram, OM ’ UV, XW || UV, UV ¼ 24 cm andXW ¼ 32 cm. If the radius of the circle is 20 cm, find thelength of MN.

U M

N

O

XV

W

12 O and P are the centres of circles. If OY ¼ 40 cm, XP ¼ 51 cm andXY ¼ 48 cm, find the distance between the centres.

X

Y

TO P

13 P and Q are the centres of circles. If DC ¼ 32 cm, PX ¼ 30 cm andXQ ¼ 12 cm, find the radius of each circle.

X

C

D

P Q

Stage 5.3

See Example 2

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Chapter 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Circle geometry

14 M and N are the centres of circles. MT ¼ 36 cm, NT ¼ 20 cm andAM ¼ 39 cm. Find:

a AB b the area of kite ANBM TM N

A

B

15 A and B are the centres of circles with the same radius. XY ¼ 40 cmand AB ¼ 96 cm. Find:

A

X

Y

Ba the radius of each circle

b the area of rhombus AXBY

15-03 Angle properties of circlesWe will now look at proofs of theorems relating to angles in a circle. These theorems can also bedemonstrated using GeoGebra, other graphing technology, or pencil-and-paper constructions.

Summary

Theorem 6: The angle at the centre of a circle is twice theangle at the circumference, standing on the same arc.

A

O

B

C

\AOB ¼ 2 3 \ACB

The arms of \ACB touch the endpoints of the arc AB, so we say that \ACB stands on the arc AB.\AOB is the angle at the centre and \ACB is the angle at the circumference. Both angles stand onthe same arc AB.

Proof:Draw radius CO and produce (extend) to point D.

Ax

O

D

B

x

C

y

y

Let \ACD ¼ x and \BCD ¼ y.

[ \ACB ¼ x þ y

OA ¼ OC ¼ OB (equal radii)

[ \OAC ¼ x (equal angles in isosceles 4OAC)

Similarly, \OBC ¼ y (equal angles in isosceles 4OBC)

\AOD ¼ xþ x ðexterior angle of 4OACÞ¼ 2x

Similarly; \BOD ¼ yþ y ðexterior angle of 4OBCÞ¼ 2y

) \AOB ¼ 2xþ 2y

¼ 2ðxþ yÞ¼ 2 3\ACB

So the angle at the centre of the circle is twice the angle at the circumference, standing on thesame arc.

Stage 5.3

Worksheet

A page of circles

MAT10MGWK10234

Puzzle sheet

Finding angles in circles

MAT10MGPS00054

Worksheet

Angles problems withalgebra

MAT10MGWK00040

5679780170194662


Example 3

Find the value of each pronumeral.

cba

O36°

x° O220°

y°Y

OBS

d°61°

Solutiona x ¼ 2 3 36 ðangle at centre is twice angle at circumferenceÞ¼ 72

b y ¼ 12

3 220 ðangle at centre is twice angle at circumferenceÞ

¼ 110c \BSY ¼ 2 3 61� ðangle at centre is twice angle at circumferenceÞ

¼ 122�

SB ¼ SY (equal radii)[\SBY ¼ \SYB ¼ d� (equal angles in isosceles 4SYB)2d þ 122 ¼ 180 ðangle sum of 4SYBÞ

2d ¼ 58

d ¼ 29

Summary

Theorem 7: The angle in a semicircle is a right angle.A

C

BO

∠ACB = 90°

Proof:

AO

C

B

\AOC ¼ 180� (a straight angle)

\ABC ¼ 12

of \AOC ðangle at centre is twice angle at circumferenceÞ

¼ 12

3 180�

¼ 90�

The angle in a semicircle is a right angle.

Stage 5.3

Video tutorial

Circle geometry

MAT10MGVT10020

568 9780170194662

Chapter 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Circle geometry

Summary

Theorem 8: Angles at the circumference of a circle, standing on thesame arc, are equal.Or put another way, angles in the same segment are equal.

AB

MN\A ¼ \B

Angles \A and \B both stand on the same arc MN.

Proof:Join radii OM and ON.

Let \MON ¼ 2x

) \A ¼ 12

3\MON ðangle at the centre is twice the angle at the circumferenceÞ

¼ 12

3 2x

¼ x

AB

N

M

2x

xO

x

Similarly;\B ¼ 12

3\MON

¼ 12

3 2x

¼ x

[ \A ¼ \B

So angles standing on the same arc are equal.

Example 4

Find b, giving reasons. N

b°

A

OV

48°

Solution\NAO ¼ 90� (angle in a semicircle)

) \VAO ¼ 90� � 48�

¼ 42�

[ b ¼ 42 (angles standing on the same arc)

Cyclic quadrilateralsA cyclic quadrilateral is a quadrilateral whose four vertices lie on thecircumference of a circle. cyclic

quadrilateral

Stage 5.3

Video tutorial

Circle geometry

MAT10MGVT10020

5699780170194662


Summary

Theorem 9: The opposite angles in a cyclic quadrilateral aresupplementary.

OA

BC

D

\A þ \C ¼ 180�\B þ \D ¼ 180�

Proof:Draw radii OA and OC.

Let \B ¼ x� and \D ¼ y�.

A

x°

y°

2y°

2x°O

B

C

D

[ Reflex \AOC ¼ 2x� (angle at centre is twice angle at thecircumference)

[ Obtuse \AOC ¼ 2y� (angle at centre is twice angle at thecircumference)

2xþ 2y ¼ 360 ðangles at a pointÞxþ y ¼ 180

[ \B þ \D ¼ 180�Hence, \A þ \C ¼ 180� (angle sum of a quadrilateral)

So the opposite angles of a cyclic quadrilateral are supplementary.

Summary

Theorem 10: The exterior angle at a vertex of a cyclic quadrilateralis equal to the interior opposite angle.

A

B

C

D

E

\CDE ¼ \B

Proof:Let \B ¼ x�.

A

x°

B

C

D

180° – x°

E

\ADC ¼ 180� � x� (opposite angles of a cyclic quadrilateral)

\CDE ¼ 180� � ð180� � x�Þ ðangles on a straight lineÞ¼ 180� 180þ x�

¼ x�

[ \ABC ¼ \CDE ¼ x�

The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

Stage 5.3

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Chapter 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Circle geometry

Example 5

Find p and q, giving reasons.

O

q°

p°

47°Solutionp ¼ 180� 47 ðopposite angles of a cyclic quadrilateralÞ¼ 133

q ¼ 2 3 133 ðangle at centre is twice angle at circumferenceÞ¼ 266

Example 6

Find c, giving reasons. E

Y NC°

S

A43°77°

Solution\ASN ¼ 77� (exterior angle of cyclic quadrilateral YESN)

c ¼ 180� 43� 77 ðangle sum of 4SANÞ¼ 60

Exercise 15-03 Angle properties of circles1 Find x, giving reasons, if O is the centre of the circle.

a b c

d e f

g h i

Ox°

Ox°

56°

O

x°

240°

Ox°

116°

Ox°

80°

O

x°

37°

O x°126° O

x°52° O

x°

Stage 5.3

Video tutorial

Circle geometry

MAT10MGVT10020

See Example 3

5719780170194662


2 Find x, giving reasons.

a b c

d e f

x° 48°x°

36°

x°30°

40°

x°

35°x°

74°

x°

3 Find the value of each pronumeral, giving reasons.

Ra b c

d e f

S

T

OO144°

y°

x°

43° Om° n°

37°

O84°

25°

h°w°

100°O

p°

37°


105°

m°

59°

92°

p°

q°

180°

x°

y°

O

a b c


106°

110°

84°

n°

w° x°

y°

a b c

Stage 5.3

See Example 4

See Example 5

See Example 6

572 9780170194662

Chapter 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Circle geometry

6 Find x, giving reasons.

a b c

d e f

Ox°

67°

O3x°

7x° O

x°

x°

117°

75°

x°

x°

88°


cba

fed

A

O

C

D

B

y°

z°x°

150°33°

A

C

D

B

Ox°

z°

y°

54°y°z°

x° 110°

A

CB

x°y°

z°

48°

A

O

CB

y°

z°x°20°

O

x°z°

y°

28°

8 Which of these quadrilaterals can also be a cyclic quadrilateral? Why?

CBA P

S R

Q

85°

48° 105°

122°

D

F

E

G

47°

118° 141°

54°

Z

W

X

Y

23°

49° 157°

131°

15-04Tangent and secant propertiesof circles

We will now look at proofs of theorems relating to tangents and secants in a circle. Thesetheorems can also be demonstrated using GeoGebra, other graphing technology, or pencil-and-paper constructions.

Stage 5.3

NSW

Worksheet

Circle geometry cards

MAT10MGWK10235

Worksheet

Circle geometry review

MAT10MGWK10236

5739780170194662


Tangents to a circleA tangent to a circle is a line that touches the circle at only one point. The point of intersection iscalled the point of contact.At any point on a circle, there is only one tangent to the circle at that point.

Summary

Theorem 11: A tangent to a circle is perpendicular to theradius drawn to the point of contact.

O

PX

T\OPT ¼ 90�

Theorem 12: The two tangents drawn to a circle from anexternal point are equal in length.

P

M

T

PT ¼ PM.

Proof:Draw radii OT and OM, and join OP.

O

PT

M

In 4PTO and 4PMO:OP is common.OT ¼ OM (equal radii)\OTP ¼ \OMP ¼ 90� (angle between tangent and radius)[ 4PTO ” 4PMO (RHS)[ PT ¼ PM (matching sides of congruent triangles)So tangents drawn from an external point are equal in length.

Alternate segment theorem

Summary

Theorem 13: The angle between a tangent and a chorddrawn to the point of contact is equal to the angle in thealternate segment.

A BP

YM

\YPB ¼ \M

(To be proved in question 1 of Exercise 15-04)AB is a tangent to the circle. The chord PY divides the circle into two segments.\M is in the alternate (opposite) segment to \YPB.

Similarly, \Y is in the alternate segment to \APM, so \APM ¼ \Y.

Stage 5.3

Worksheet

A page of circles

MAT10MGWK10234

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Chapter 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Circle geometry

Example 7

Find the value of each pronumeral, giving reasons.

A

O

C

B40°

x°

85 mm

d

y°

x°65°

37°T

a°

I

K

Aa b cR

31°

Solutiona \KTR ¼ 90� (angle between a tangent and a radius)

aþ 31 ¼ 90

a ¼ 59b d ¼ 85 (tangents from an external point)\ABO ¼ \ACO ¼ 90� (angle between a tangent and a radius)

) xþ 90þ 40þ 90 ¼ 360 ðangle sum of a quadrilateralÞxþ 220 ¼ 360

x ¼ 140c x ¼ 65 (angle in alternate segment)

y ¼ 37 (angle in alternate segment)

Summary

Theorem 14: When two circles touch,their centres and the point of contactare collinear.

A

B

TQP

A

B

P QTThe circles have a common tangent AB.P, T and Q lie on a straight line (to beproved in question 2 of Exercise 15-04).

Products of interceptsWhen a line crosses another line or curve at two points, the distance between the points is calledan intercept.

Summary

Theorem 15: The products of the intercepts of two intersectingchords are equal.

D

A

B

C

YAY 3 YB ¼ CY 3 YD

AY, YB, CY and YD are called intercepts.

Stage 5.3

Video tutorial

Circle geometry

MAT10MGVT10020

5759780170194662


Proof:Draw AD and CB. D

A

B

C

Y

In 4ADY and 4CBY:\D ¼ \B (angles in the same segment)\A ¼ \C (angles in the same segment)

[ 4ADY ||| 4CBY (equiangular)

)AYCY¼ YD

YB(matching sides of similar triangles)

[ AY 3 YB ¼ CY 3 YD

Therefore, the products of intercepts of two intersecting chords are equal.

Summary

C

AB

D

Y

Theorem 16: The product of the intercepts of two intersectingsecants to a circle from an external point are equal.YA 3 YB ¼ YC 3 YD

AB

T

P

Theorem 17: The square of a tangent to a circle from anexternal point equals the product of the intercepts of anysecant from the point.

PT2 ¼ PB 3 PA

(To be proved in question 3 of Exercise 15-04.)

Example 8

Find x, giving reasons.

cba

C

D

B

A

Y

x cm

15 cm

20 cm 9 cm

A

CD

B

Y

10 cm8 cm

6 cm

x cm

A

T

B

P

48 cm16 cm

x cm

Solutiona x 3 15 ¼ 9 3 20 ðintercepts of intersecting chordsÞ

15x ¼ 180

x ¼ 12b 6 3 ð6þ xÞ ¼ 8 3 18 ðintercepts of intersecting secantsÞ

36þ 6x ¼ 144

6x ¼ 108

x ¼ 18

Stage 5.3

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Chapter 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Circle geometry

c x2 ¼ 16 3 ð16þ 48Þ ðtangent� secant theoremÞ¼ 1024

x ¼ffiffiffiffiffiffiffiffiffiffi

1024p

¼ 32

Exercise 15-04 Tangent and secant properties of circles1 AB is a tangent to a circle at P, with centre O.

A BP

M

YOa If \YPB is an angle between a tangent and chord, what is theangle in the alternate segment?

b If \YPB ¼ x�, why is \OPY ¼ 90� � x�?

c Show that \POY ¼ 2x�.

d Hence show that \PMY ¼ \YPB.

2 Two circles with centres P and Q touch at T witha common tangent AB.

T

A

B

P Q TPQ

A

B

a Explain why \ATP and \ATQ are right angles.

b Hence, show that P, T and Q are on thesame line.

T

BP

A

3 PT is a tangent and PA is a secant.a Prove that 4PTA ||| 4PBT.

b Hence, show that PTPB¼ PA

PTand PT2 ¼ PB 3 PA.


ba

dc

a° 34°

b

75

72

e°46°g°

67°

Stage 5.3

See Example 7

5779780170194662


5 Find the value of n.

cba4

512n

4

10n

(n + 3)21

6

n n

fed8

516 n

9

16

n 5

6

n

6 ABC is a right-angled triangle with an internal circle such that eachside is a tangent to the circle. R, S and T are the respective pointsof contact of each of the tangents. If SC ¼ 30 cm, RB ¼ 5 cmand AT ¼ x, find the value of x. T

R

SB C

A

7 Two circles with centres P and Q and radii 18 cm and8 cm respectively, touch each other externally as shown.Find the length of:

AB

QP

Xa XP b AB

15-05 Proofs using circle theorems

Example 9

PT is a tangent to a circle, and PA ¼ PB.

A

PT

B

Prove that PT || AB.

Solution\TPB ¼ \A (angle in alternate segment)\A ¼ \B (equal angles of isosceles 4PAB)[ \TPB ¼ \B

[ PT || AB (alternate angles are equal)

Stage 5.3

See Example 8

NSW

578 9780170194662

Chapter 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Circle geometry

Exercise 15-05 Proofs using circle theorems1 PQ and RS are two chords of a circle intersecting at the point Y. Prove

that the triangles PYR and SYQ are similar and hence PY 3 YQ ¼ RY 3 YS.R

P

Q

S

Y

2 ABCD is a cyclic quadrilateral. Side AB is produced to E so that AECD

is a parallelogram. Prove that 4CBE is an isosceles triangle.

A B E

CD

3 O is the centre of two concentric circles. DEFG is a straight line. Provethat DE ¼ FG.

D GE

O

F

4 HT is a tangent and parallel to IP. HJP is a straight line.Prove that \HIP ¼ \HJI.

HT

P

J

I

5 UVW is an isosceles triangle, where UV ¼ UW. A circle withcentre O is drawn on one of the equal sides. Prove that thecircle bisects the base of the triangle.

U

X WV

O

6 PQRS is a cyclic quadrilateral. PR bisects \QRS. Prove that 4SPQ isisosceles.

QP

RS

7 Two circles with centres P and Q intersect at X and Y. AB isparallel to PQ, the interval joining the centres of the circles.

Prove that PQ ¼ 12

AB.

X

Y BA

P Q

8 Two circles touch externally at T. XY is the common tangent.PTQ and RTS are straight lines. Prove that PR || SQ.

X

Y

TP

R

Q

S

Stage 5.3

See Example 9

5799780170194662


measurement and geometrycircle geometry...

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