Chapter 15 Circle Geometry

1 5

Measurement and geometry

Circle geometry Optional Stage 5.3 topic The ancient Greek philosopher, Thales of Miletus (624 546 BCE), is believed to have been the first mathematician and scientist to attempt to explain things by reasoning and deduction rather than by resorting to mythology and superstition. He lived before Pythagoras and proved the first theorem in mathematics: that any angle drawn on the circumference of a semicircle is a right angle. This rule is often called Thales’ theorem.



NEW CENTURY MATHS ADVANCED for the

A u s tr a l i an C u r ri c u l um

10

þ10A

s i l l E n o R / m o c . k c o t s r e t t u h S

n Chapter outline 15-01 Parts of a circle* 15-02 Chord properties of circles* 15-03 Angle properties of circles* 15-04 Tangent and secant properties of circles* 15-05 Proofs using circle theorems* *STAGE 5.3

n Wordbank Proficiency strands U C

collinear points Points that lie on a straight line. converse A rule or statement turned back-to-front; the reverse statement

U

F

R

C

U

F

R

C

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

R

C

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

R

C

perpendicular bisector An interval or line that bisects another interval at right angles

U

F

U

F

PS

A line that intersects intersects a curve at two points secant A opposite site an angle and ‘hold’ its subtend an angle To sit oppo arms A line that touches a curve at one point but tangent A does not cross it

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

n In this chapter you will: • (STAGE (STAGE 5.3) Prove Prove and apply apply angle and and chord proper properties ties of circle circless • (STAG (STAGE E 5.3) iden identify tify the the parts parts of of a circle circle • (STAG (STAGE E 5.3) prove chord chord and angle angle properties properties of circles circles and use them them to find unknown unknown lengths and angles in circles • (STAG (STAGE E 5.3) prove tangent tangent and secant secant properti properties es of circles circles and use them to find find unknown lengths and angles in circles

SkillCheck Worksheet

1

StartUp assignment 15

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

MAT10MGWK10232

b

c

m 6c cm

6 cm cm

10 cm 10 cm 10 cm

d

2

e

f

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

b

15

30

21

20

c

12

14 15

10

d

9

2.5

e

4 30°

3 3

4.5

f 3

55°

40° 6

2 110°

30° 95°

30°

8

3.75

558

20

9780170194662


A u st ra l i a n C u rr i cu lu m

10 10A

þ

Stage 5.3

15-01 Parts of a circle

Worksheet

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

A tangent is is a line touching the circle at one point. A secant is is a line that cuts the circle at two points.

Parts of a circle MAT10MGWK10233

radius n t g e n t a

centre

circumference secant

A semicircle is half of a circle.

A quadrant is is quarter of a circle.

semicircle

Quadrant

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

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

e r e a m e t d i a

segment a r c

Exercise 15-01 Parts of a circle 1

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

b c a

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Stage 5.3

2

segment in a circle. circle. a Draw a sector and a segment difference between a sector and a segment? segment? b What is the difference

3

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

4

Which parts parts of a circle circle are labelled a and b? Select the correct answer A, B , C or D . A diameter and segment C diameter and sector

a b

B radius and segment D radius and sector

5

Write the correct word to match each each description. description. centre of a circle to its side a the distance from the centre b quarter of a circle c line that touches the outside of a circle once centre d an interval from one side of a circle to the other side, through the circle’s centre side of the circle to the other side, side, not through the centre centre e an interval from one side circle f part of the circumference of a circle circle formed by two radii and and an arc g the area inside a circle distance around a circle h the complete distance inside a circle formed formed by a chord chord and an arc arc i the area inside

6

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

c a

b

d

Technology Perpe Perpendicul ndicular ar to a chord chord 1 Click View and Axes in GeoGebra.

using Circle with centre through point. 2 Draw a circle using on the circumference of the circle. 3 Insert one new point on 4 Use Interval between two points to join BC . 5 Click Midpoint or Centre

and chord BC.

and A and the midpoint of chord BC . 6 Select Perpendicular line BC and to the midpoint of BC and repeat for point B to 7 Select Distance or Length and measure C to the midpoint of BC BC . Use the Move tool to manipulate points B and C . What do you notice?

560

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10 10A

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Stage 5.3

15-02 Chord properties of circles

Worksheet

Equal chords

A page of circles

When 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 the arc or chord.

MAT10MGWK10234

P

A O B

X Y

The arc AB subtends the angle at O .

The chord XY subtends subtends the angle at P.

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

Summary C

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

D O B

A

If AB = CD, ∠ AOB = ∠COD.

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

4 ¼ ¼ ¼ 4

4

¼

4

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

C

Y

O A

X

D B

If AB = CD, then OX = OY .

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Stage 5.3

Proof: Draw radii OA , OB , OC and and OD. In AOX and and COY : (equal radii) OA OC \ A \C (matching angles of congruent AOB and COD proved in Theorem 1) \OXA \OYC 90 (given) [ AOX ” COY (AAS) [ OX OY (matching sides of congruent triangles) So equal chords are equidistant from the centre of a circle.

4 ¼ ¼

4

C

4

4

¼

¼

4

Y

D

O

4

B

X

A

¼

Perpendicular bisectors of chords Summary Theorem 3: The perpendicular from the centre of a circle to a chord bisects the chord. The converse of this is also true. The line from the centre of a circle to the midpoint of a chord is perpendicular to the chord.

O

A

B

X

The word ‘converse’ means to ‘turn around’ or ‘reverse’. So the converse of a theorem is the theorem written ‘back-to-front’. Proof: Draw radii OA and OB. In AOX and and BOX : (equal radii) OA OB A is common. OX is \OXA \OXB 90 (OX ’ AB) [ AOX ” BOX (RHS) (matching sides of congruent triangles) [ AX BX So the perpendicular from the centre of a circle to a chord bisects the chord.

4 ¼

4

4

¼

¼

4

O X

¼

B

Summary Theorem 4: The perpendicular bisector of a chord in a circle passes through the centre of the circle.

O A X

562

B

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Proof: The perpendicular from O to AB bisects AB . (pro (p rove ved d in Th Theo eore rem m 3) [ OX is is the perpendicular bisector of AB. [ The perpendicular bisector of AB AB passes through the centre O .

10 10A

þ

Stage 5.3

Common chords Summary A

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

P B

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

A

In

4OAP and and 4OBP : OA ¼ OB (equal radii) (equal radii) PA ¼ PB

O

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

4

4 ¼

4 ¼

4

¼

¼

M

P

B

4

4

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. PQ a If PQ

¼ 35 cm, find PT . Q

9780170194662

¼

A

O P

OP 16 cm, find AB . b If OP

T

16 P

O 34 B

563


Stage 5.3

Solution

¼¼ 12 ð PQÞ ðthe perpendicular from the centre bisects the chord Þ ¼ 12 35 ¼ 17 5 cm by Pythagoras’ theorem b PB2 ¼ 342  162 ¼ 900 p ffiffiffiffiffiffiffiffi ffi PB ¼ 900 ¼ 30 cm ) AB ¼ 2 PB ðthe perpendicular from the centre bisects the chord Þ ¼ 60 cm a

PT

3

:

3

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 the distance between their centres.

A 13 cm X

15 cm Y

B AB = 24 cm

Solution We need to find XY.

A

Let XY and and AB cross at M . (common chord of circles)

XY ’ AB

¼¼ 12 AB ¼ 12 24 ¼ 12 cm XM 2 ¼ 132  122 ¼ 25 p ¼ ffiffiffiffiffi XM ¼ 25 ¼ 5 cm YM 2 ¼ 152  122 ¼ 81 p ffiffiffiffiffi ¼ 81 YM ¼ ¼ 9 cm ¼ 5 þ 9 ) XY ¼ ¼ 14 cm AM

X

M

Y

B

3

3

by Pythagoras’ theorem

The distance between the centres is 14 cm.

564

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10 10A

þ

Stage 5.3

Exercise 15-02 Chord properties of circles D

subtending equal angles at the 1 The converse of Theorem 1 is: ‘Chords subtending centre of a circle have equal lengths’. a Prove that AOB ” COD, given that \ AOB b Hence, prove that AB CD.

4

4 ¼

C O

¼ \COD.

A B ∠ AOB = ∠ COD C

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

Y O

and therefore AX CY. a Prove that OXA ” OYC and b Prove that OXB ” OYD and therefore BX DY. c Hence show that AB CD.

4 4

4 4 ¼

¼ ¼

4

4 a b c d

4

D

A X

B OX = OY

3 The converse of Theorem 3 is: ‘The line from the centre of a circle to the midpoint of a chord is perpendicular to the chord.’ a Prove that OXA ” OXB and therefore \ OXA b Prove that OX ’ AB.

Example 1 See Example

O

¼ \OXB.

A

X

B

AX = BX

Draw a circle and construct construct any two non-parallel non-parallel chords. Bisect these these chords. chords. Through what point must must each perpendicular perpendicular bisector pass? (Use (Use Theorem 4.) Explain why the point of intersection between the two perpendicular perpendicular bisectors is the centre of the circle. C

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

Y A

O B

X

The full statement on this theorem is: ‘Given any three non-collinear* points, the point of intersection of the perpendicular bisectors of any two sides of the triangle formed by the three points is the centre of the circle through all three points. (*Non-collinear means not on on the same straight line.) Prove this theorem as follows: and therefore OA OC a prove that AOY ” COY and and therefore OA OB b prove that AOX ” BOX and that the circle with centre O and radius OA also passes through B and C . c hence prove that

4 4

4 4

¼ ¼

S

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

A R

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O

565


Stage 5.3

7 In the diagram, AB find AB.

’ OP .

If OB OB

¼ 32.5 cm and OP ¼ 12.5 cm,

A O P B

theorem you use. 8 In each problem, state which chord theorem a OU OV b GF 12 m Find the length of DE AB 9 m DE . Find the length of UC UC .

¼ ¼

¼

c UV

¼ VW \VOW ¼ 64

Find the size of \UVO.

E D

V

B

O U

V

A

C

U

O

D

¼ ¼

G

O

e OM ’ XY XY 40 cm OX 29 cm Find OM .

f DE ’ ON ON 18 m DE 36 m Find the radius.

¼ ¼

¼ ¼

X O T

W

F

d OT ’ PQ OQ 17 mm OT 8 mm Find PQ .

P



O

N

D

M

E

O

Q Y

radius of a circle in which a chord of length 96 cm is 20 cm from the centre. centre. 9 Find the radius 10 On a clear day, you can see 11.5 km in any direction from a particular lookout. If a straight road is 6.9 km from the lookout, what length of the road can be seen from the lookout?

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

¼

¼

O N

X

M

U Example 2 See Example

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

¼

¼

W

¼

V

X O

P

T

Y

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

¼

¼

¼ 30 cm and

C P

X

Q

D

566

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and N are are the centres of circles. MT 36 cm, NT 14 M and AM 39 cm. Find: area of kite kite ANBM a AB b the area

¼

¼

¼ 20 cm and

10 10A

þ

A M

Stage 5.3 N

T

B

15 A and B are the centres of circles with the same radius. XY and AB 96 cm. Find: circle a the radius of each circle rhombus AXBY b the area of rhombus

¼

¼ 40 cm

X A

B Y

Worksheet

15-03 Angle properties of circles

A page of circles MAT10MGWK10234

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

Puzzle sheet Finding angles in circles MAT10MGPS00054

Summary

Worksheet

C

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

¼2

3

Angles problems with algebra

O

\ ACB

MAT10MGWK00040

A B

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 on the same arc AB. Proof: Draw radius CO and produce (extend) to point D.

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 ) ðexterior angle of 4 4OAC Þ \ AOD ¼ x þ x ¼ 2 x 4OBC Þ Similarly \ BOD ¼ y þ y ðexterior angle of 4 ¼ 2 y ) \ AOB ¼ 2 x þ 2 y ¼ 2ð x þ yÞ ¼ 2 \ ACB

C x y

Let \ ACD

O

[

x A

D y

[

B

;

3

So the angle at the centre of the circle is twice the angle at the circumference, standing on the same arc. 9780170194662

567


Stage 5.3 Video tutorial Circle geometry

Example

3

Find the value of each pronumeral. a

MAT10MGVT10020

b 36° O

c 220° O

x°

B d °

O

S 61°

y° Y

Solution

¼ 2 36 ðangle at centre is twice angle at circumferenceÞ ¼ 72 1 220 ðangle at centre is twice angle at circumference Þ y¼ 2 ¼ 110 ¼ 2 61 ðangle at centre is twice angle at circumferenceÞ \ BSY ¼ ¼ 122 SB ¼ SY (equal radii) (equal angles in isosceles 4SYB) \SBY ¼ \SYB ¼ d 4SYBÞ 2d þ 122 ¼ 180 ðangle sum of 4 ¼ 58 2d ¼ ¼ 29 d ¼

a x

3

b

3

c

3



[

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

C A O

B

∠ ACB = 90°

B

Proof:

¼ 180 (a straight angle) ¼ 12 of \ AOC ð ðangle at centre is twice angle at circumference Þ \ ABC ¼ ¼ 12 180 ¼ 90

\ AOC



C A

O

3

The angle in a semicircle is a right angle.

568

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10 10A

þ

Stage 5.3

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

B A

¼ \ B

N M

Angles \ A and \ B both stand on the same arc MN . Proof: Join radii OM and and ON.

Let \ MON 2 x 1 3 \ MON ) \ A 2 1 3 2 x 2

¼

¼ ¼ ¼ x

Similarly \ B ;

[

\ A

1 2 1 2

¼ ¼ ¼ x

ðangle at the centre is twice the angle at the circumference Þ B

A x

x O

2x

3 \ MON

N

M 3

2 x

¼ \ B

So angles standing on the same arc are equal.

Example

4

Video tutorial Circle geometry

N

Find b , giving reasons.

MAT10MGVT10020

b° 48° A V

Solution

O

¼ 90 (angle in a semicircle)   ) \VAO ¼ 90  48 ¼ 42 b ¼ 42 (ang (a ngle less st stan andi ding ng on th thee sa same me ar arc) c)

\ NAO



[

Cyclic quadrilaterals A cyclic quadrilateral is a quadrilateral whose four vertices lie on the circumference of a circle.

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cyclic quadrilateral

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Stage 5.3

Summary Theorem 9: The opposite angles in a cyclic quadrilateral are supplementary.

þ \C ¼ 180 \ B þ \ D ¼ 180 \ A

B C



A

O



D

Proof: Draw radii OA and OC. Let \ B x and \ D y.

¼

[ Reflex

B x°

¼

\ AOC

¼ 2 x

[ Obtuse \ AOC

(angle at centre is twice angle at the circumference) (angle at centre is twice angle at the circumference)



¼ 2 y



C 2 y° O 2x°

A

y°

2 x

þ 2 y ¼ 360 ðangles at a pointÞ x þ y ¼ 180 \ B þ \ D ¼ 180 Hence, \ A þ \C ¼ 180 (angle sum of a quadrilateral quadrilateral))

D



[



So the opposite angles of a cyclic quadrilateral are supplementary.

Summary Theorem 10: The exterior angle at a vertex of a cyclic quadrilateral is equal to the interior opposite angle. \CDE

¼ \ B

B C

A D E

Proof:

Let \ B

¼ x . (opposite angles of a cyclic quadrilatera quadrilateral) l) \ ADC ¼ 180  x    \CDE ¼ 180  ð180  x Þ ðangles on a straight line Þ ¼ 180  180 þ x ¼ x \ ABC ¼ \CDE ¼ x 

[

B





A

180° – x°



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

570

C

x°

D E

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Example


10 10A

þ

Stage 5.3

5 q°

Find p and q , giving reasons. 47°

Solution

O

¼ 180  47 ðopposite angles of a cyclic quadrilateral Þ ¼ 133 q ¼ 2 133 ðangle at centre is twice angle at circumference Þ ¼ 266

p

p°

3

Example

Video tutorial

6

Circle geometry

Find c, giving reasons.

E

MAT10MGVT10020

Solution

S

¼ 77 (exterior angle of cyclic quadrilateral YESN ) 4SAN Þ c ¼ 180  43  77 ðangle sum of 4 ¼ 60

\ ASN



7 7°

Y

C°

N

43°

A

Exercise 15-03 Angle properties of circles 1 Find x , giving reasons, if O is the centre of the circle. a

b

x° O


c 56° O x°

240° O x°

d

e

f

x° O

O

116°

h 126°

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O

80°

g O

37°

x°

x°

x°

i 52°

x° O

x° O

571


Stage 5.3

2 Find x , giving reasons. a


b x°

c

x°

x°

48°

40°

36°

d

e

f

35°

x° x°

x°

3

30°

74°

Find the value of each each pronumeral, pronumeral, giving reasons. a

b

S

c

37°

R

T

144°

O y°

O

4 3 °

m° O n°

x°

d

e

f

p°

25°

w°

O

O 84°

37° 100°


4

h°


b

105°

59°

c

p°

x° O 180°

m° Example 6 See Example

5

92°

q°

y°


b

c

x° 110°

w° 106°

n° y°

572

84°

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10 10A

6 Find x , giving reasons. a

Stage 5.3

b

c

67° O

7x°

3x°

x°

d

x°

e

f

x°

x°

88°

75°

x°

Find the value of each pronumeral, pronumeral, giving giving reasons. reasons. a

A

b

D O

33° 150° B

z°

z°O x°

y° B

y° C

e

f

A

y°

y°

48°

y°

D x° 110°

A

B

z° y°

54°

C

d

c

A

x°

8

O

O

117°

7

þ

x°

z° x° C

B

20°

O x°

z° 28° z°

O

C

Which of these quadrilateral quadrilateralss can also be a cyclic quadrilateral? quadrilateral? Why? Why? B D

P

A

85°

54° 122°

S

48°

105°

R

W

C

47°

E

23°

Q

131° X

G

118°

141°

F Z

49°

157°

Y

Tangent and secant properties 15-04 of circles

NSW Worksheet

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

Circle geometry cards MAT10MGWK10235 Worksheet Circle geometry review MAT10MGWK10236

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Stage 5.3 Worksheet A page of circles MAT10MGWK10234

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

Summary X

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

¼ 90



T

O

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

P

T P

¼ PM .

M

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

T

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

4

¼

4

4

¼

¼

4

P

O M

¼

Alternate segment theorem Summary Theorem 13: The angle between a tangent and a chord drawn to the point of contact is equal to the angle in the alternate segment. \YPB

M

A

¼ \ M

Y

P

B

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

¼

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Example


10 10A

þ

Stage 5.3

7

Video tutorial

Find the value of each pronumeral, giving reasons. a

b

A K

R 31°

8 5 mm

B

c

A

MAT10MGVT10020

40°

a° T

Circle geometry

O

x°

37°

d

x° 65°

y°

C I

Solution a

b

¼ 90 (angle between a tangent and a radius) a þ 31 ¼ 90 a ¼ 59 d ¼ 85 (tan (t ange gent ntss fr from om an ex exte tern rnal al po poin int) t) (angle between a tangent and a radius) \ ABO ¼ \ ACO ¼ 90 ) x þ 90 þ 40 þ 90 ¼ 360 ðangle sum of a quadrilateral Þ x þ 220 ¼ 360 x ¼ 140 x ¼ 65 (angle in alternate segment) y ¼ 37 (angle in alternate segment) \ KTR





c

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

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

A

A

P

T

Q

T

P Q

B

B

Products of intercepts When a line crosses another line or curve at two points, the distance between the points is called an intercept .

Summary D

Theorem 15: The products of the intercepts of two intersecting chords are equal. AY 3 YB

¼ CY

3

B Y

YD A

C

and YD are called intercepts. AY , YB , CY and 9780170194662

575


Stage 5.3

Proof: AD and CB . Draw AD In ADY and and CBY : \ D \ B (angles in the same segment) (angles in the same segment) \ A \C ||| CBY (equiangular) [ ADY |||

D

4 4 ¼ ¼ 4 4 AY YD ¼ (matching sides of similar triangles) ) CY YB AY YB ¼ CY YD [

3

B

Y A C

3

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

Summary Theorem 16: The product of the intercepts of two intersecting secants to a circle from an external point are equal. YA 3 YB

¼ YC

3

C D Y

YD B A

Theorem 17: The square of a tangent to a circle from an external point equals the product of the intercepts of any secant from the point. PT 2

¼ PB

T P

3 PA

B A

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

Example

8

Find x , giving reasons. A

a

b

A

10 c m m

c

B 8 c m m

15 cm

C

20 cm Y 9 cm D x cm

C

m

x c

D

6 c m

A

4 8 8 c m m

B 16 c m m P

Y m

x c

T

B

Solution a x 3 15 15 x

¼ 9 20 ðintercepts of intersecting chords Þ ¼ 180 x ¼ 12 ð6 þ xÞ ¼ 8 18 ðintercepts of intersecting secants Þ 36 þ 6 x ¼ 144 6 x ¼ 108 x ¼ 18

b 63

576

3

3

9780170194662


c x2

¼ 16 ð16 þ 48Þ ¼ 1024 p ffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffiffi ffi x ¼ 1024 ¼ 32 3


10 10A

þ

ðtangent  secant theoremÞ

Stage 5.3

Exercise 15-04 Tangent and secant properties of circles 1

AB is a tangent to a circle at P , with centre O .

M

a If \YPB is an angle between a tangent and chord, what is the angle in the alternate segment? b If \YPB x, why is \ OPY 90 c Show that \ POY 2 x. show that \ PMY \YPB. d Hence show

¼

2

¼

¼

O

 x ? 

A

¼

Two circles with centres P and and Q touch at T with with a common tangent AB. and \ ATQ are right angles. a Explain why \ ATP and show that P , T and and Q are on the b Hence, show same line. is a tangent and PA is a secant. PT is a Prove that PTA ||| PBT . show that PT PA and PT 2 b Hence, show PB PT

4

B

P

A

P

A

Q

T

P

B

3

Y

4 ¼

T

Q

B T

¼ PB

.

3 PA

P B A

4

Find the value of each pronumeral, pronumeral, giving giving reasons. reasons. a


b a°

75

34° b

c

72

d 46°

e°

g ° 67°

9780170194662

577


Stage 5.3

5

Find the value of n n . a


b

4

10

12 n

5

c n

(n + 3)

n

n 6

21

4

d

n

e

f

5

8

n

16 n

16

6

9

5

6

A

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

¼

¼

¼

T R B

7

Two circles with centres P and and Q and radii 18 cm and 8 cm respectively, touch each other externally as shown. Find the length of: a XP

NSW

b AB

S

C

A B X Q

P

15-05 Proofs using circle theorems Example

9

PT is is a tangent to a circle, and PA

P

¼ PB.

T

Prove that PT || || AB .

A

Solution \TPB

|| AB [ PT ||

578

(angle in alternate segment) (equal angles of isosceles PAB)

¼ \ B \ TPB ¼ \ B

\ A [

¼ \ A

B

4

(alternate angles are equal)

9780170194662



10 10A

þ

Stage 5.3

Exercise 15-05 Proofs using circle theorems 1

are two chords of a circle intersecting at the point Y . Prove PQ and RS are that the triangles PYR and SYQ are similar and hence PY 3 YQ RY 3 YS .

Q

R

¼


Y P S A

so that AECD 2 ABCD is a cyclic quadrilateral. Side AB is produced to E so is a parallelogram. Prove that CBE is is an isosceles triangle.

B

E

4

D

3

C

is a straight line. Prove O is the centre of two concentric circles. DEFG is that DE FG .

¼

O D

4

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

E

H

T

¼

J P

I

5

UVW is is an isosceles triangle, where UV UW . A circle with centre O is drawn on one of the equal sides. Prove that the

¼

U

O

circle bisects the base of the triangle. V

is a cyclic quadrilateral. PR bisects \QRS . Prove that 6 PQRS is isosceles.

W

X

4SPQ is

Q

P

S

7

Two circles with centres P and and Q intersect at X and and Y . AB is parallel to PQ , the interval joining the centres of the circles. Prove that PQ 1 AB. 2

R X

¼

P

Q

A

8

G

F

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

B

Y X S

P T

Q R 9780170194662

Y

579

Chapter 15 Circle Geometry

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