mq10 vic ch 10a - wordpress.com · tangents parts of a circle ... since the radius og is...

1Jesse is trying to sail his yacht solo around the world. However, the yacht begins to take on water and so Jesse radios for help, giving his coordinates as (26°S, 17°W). A military vessel at (30°S, 17°W) receives Jesse’s mayday signal and heads to the area at a speed of 30 knots. Simultaneously, a search plane is dispatched from the city of Dakar (14°N, 17°W) at a speed of 750 km/h. Will the military vessel or the plane reach Jesse first?

In this chapter we will be studying the circle and circles that are formed on the surface of the Earth. We will be able to solve this problem by using these circles to calculate the distance of the plane and the military vessel from Jesse.

0Circle geometry

MQ10 VIC ch 10A Page 359 Tuesday, November 20, 2001 2:31 PM

360 M a t h s Q u e s t 1 0 f o r V i c t o r i a

Intersecting chords, secants and tangents

Parts of a circleRecall the following definitions for various parts of a circle.

Part (name) Description Diagram

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

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

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

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

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

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

O

O

O

O

O


C h a p t e r 1 0 C i r c l e g e o m e t r y 361

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

Arc A portion of the circumference.

Tangent A straight line that intersects with (that is, touches) the circumference at one point only.

Secant A chord extended beyond the circumference on one side.

Constructing a tangentThere are a number of ways to construct a tangent to a circle. One of the techniques is outlined below.1 Draw a circle of radius 5 cm and centre O.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 Investigate another technique for constructing a tangent

to a circle.8 Write a set of instructions for this method of constructing

a tangent.

Part (name) Description Diagram

O

O

O

O

O

O

O QP

MQ10 VIC ch 10A 61 Tuesday, November 20, 2001 2:31 PM


Intersecting chords

The results of the above activity can be generalised for any circle as follows.

Theorem 1If 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 thesame.

PX × XQ = RX × XS or a × b = c × d

Intersecting chordsIn the diagram below chords PQ and RS intersect at X.

1 Measure lengths PX, XQ, RX and XS and complete the table below.

2 Calculate the following: PX × XQ and RX × XS3 What do you notice about the results in step 2 above?4 Draw another circle and a pair of intersecting chords and repeat steps 1 to 3

above.

Intersecting chords

X

Q

S

P

RLine segment PX XQ RX XS

Length

X

a

c

d

b

Q

S

P

R

Find the value of the pronumeral.

THINK WRITEChords 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 × XB = CX × XD

Identify the lengths of the line segments.

AX = 4, XB = m, CX = 6, XD = 5

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

4m = 6 × 5

m =

= 7.5

1

2

3

304

------

1WORKEDExample

B

m64 5 D

A

C

X



Intersecting secants

Your observation from the activity above can be generalised as follows.

Theorem 2If two secants intersect outside the circle as shown, then the following relationship is always true: AX × XB = CX × XD or a × b = c × d.

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

Measure lengths AX, XB, CX and XD and calculate the products AX × XB and CX × XD. What do you notice?

Cabri Geometry

Intersectingsecants

B

D

A

C

X

B

D

A

C

Xdb

a

c


THINK WRITESecants CX and AX intersect outside the circle at X. Write the rule connecting the lengths of CX, XD, AX and XB.

CX × XD = AX × XB

State the length of the required line segments.

CX = y + 6 XD = 6 AX = 7 + 5 XB = 7 = 12

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

(y + 6) × 6 = 12 × 76y + 36 = 84

6y = 48 y = 8

1

2

3

2WORKEDExample

B

DA

C

X

6

y

5

7



A special case of theorem 2 applies when one of the lines drawn from an external pointis a tangent rather than a secant.

Theorem 3If a tangent and a secant intersect as shown, the following relationship is always true:AX × XB = (XT)2 or a × b = c2.

Intersecting tangentsIn the diagram at right, tangents AC and BC intersect at C and AC = BC.

Theorem 4If two tangents meet outside a circle, then the lengths from the external point to where they meet the circle are equal.

c T

B

A

Xb

a


THINK WRITESecant AX and tangent TX intersect at X. Write the rule connecting the lengths of AX, XB and XT.

AX × XB = (XT)2

State the values of AX, XB and XT. AX = m + 4, XB = 4, XT = 8Substitute the values of AX, XB and XT into the equation and solve for m.

(m + 4) × 4 = 82

4m + 16 = 64 4m = 48 m = 12

1

23

3WORKEDExample

8T

B

A

X

m4

B

A

CIntersecting tangents


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

AC = BC

State the lengths of BC and AC. AC = m, BC = 3Substitute the required lengths into the equation to find the value of m.

m = 3

1

23

4WORKEDExample

A

m

3

B

C



Chords and radiiIn the diagram at right, chord AB and the radius OCintersect at X at 90°; that is, ∠OXB = 90°.

OC bisects the chord AB, that is, AX = XB.

Theorem 5If 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 6Chords 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.

It is important that we are able to prove these theorems. To prove a theorem is true, weuse previously established theorems. We begin by stating the aim of the proof and usethe given information to establish the result we are aiming for, giving a reason for eachstep. If the reason is that the information is given to us we simply write ‘data’ as thereason.

We may need to construct new lines in the diagram and you should be aware manyproofs are completed using congruent triangles.

Cabri Geometry

ChordsBXA

C

O

B

N

AC

O

M P

RD

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

THINK WRITESince the radius OG is perpendicular to the chord AB, the radius bisects the chord.

AE = EB

State the lengths of AE and EB. AE = m, EB = 3

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

m = 3

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

State the lengths of OE and OF. OE = n; OF = 2.5

Substitute the lengths into the equation to find the value of n.

n = 2.5

1

2

3

4

5

6

5WORKEDExample

B

C

A

O

mn

G

F

E

D2.5

3

H



Prove the result that ‘If a radius meets a chord at right angles, the radius bisects the chord’.

THINK WRITEState the aim. To show OC bisects AB we need to show AX = BX.

Aim: to show AX = BX.

Construct lines OA and OB.Two triangles are formed which, if congruent, will show AX = BX.

List the measurements in the two triangles that are equal giving a reason in each case.

∠OXA = ∠OXB = 90° (data)OX is a common side.OA = OB (radii of the circle)

In each right-angled triangle, one side is common and hence equal in length and the hypotenuses are equal since they are the radii of the circle. Therefore the triangles are congruent.

Hence, the triangles are congruent. (RHS — right angle, hypotenuse, side.)That is, ∆OXA ∆OXB.

AX and BX are corresponding sides of congruent triangles.

AX = BX (corresponding sides of congruent triangles)Since AX = BX, radius OC bisects the chord AB

1

2

BXA

C

O3

4

5

=~

6

6WORKEDExample

BXA

C

O

remember1.

PX × XQ = RX × XS

3.

AX × XB = (XT)2

5.

2.

AX × XB = CX × XD

4.

AC = BC

6.

X

Q

S

P

R

c T

B

A

Xb

a

BXA

C

O If OC ⊥ AB, AX = XB.

B

D

A

C

X

B

A

C

B

N

AC

O

M P

RD

If MN = PR, then OC = OD.

remember



Intersecting chords, secants and tangents

1 Find the value of the pronumeral in each of the following.



4 Find the value of the pronumerals in each of the following.


10AWWORKEDORKEDEExample

1

A D

BC

X

m 4

86

A

D

B

C

Xm

629

A

D B

C

X

m

m

49

a b c

WWORKEDORKEDEExample

2

n

42

3m 4.5

7 6

n

48

3

a b c


3

w

45

x5

15y

412

a b c


4

x

5

m

7

y

x3.1

2.5

a b c


5

Ox

2.83.3

O

x

2.52.55.6O

xm

O

a b

c d



6

In which of the following figures is it impossible to find the value of m throughsolving a linear equation?

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

8 Prove the result: ‘If a radius bisects a chord, then the radius meets the chord at rightangles’.

9 Prove the result: ‘Chords of equal length are equidistant from the centre’.

10 Prove the converse of the previous result; that is, ‘Chords that are an equal distancefrom the centre are equal in length’.

Angles in a circle1 Draw a circle of radius 10 cm.

a Mark 5 points A, B, C, D and E on the circumference as shown below.

b Join points O, E, D and C by straight lines to points A and B.

c Measure ∠AEB, ∠ADB and ∠ACB. Comment on the results obtained.

d Measure ∠AOB. What is the relationship between this angle and the angles discussed in part c above?

2 Draw a circle and draw a diameter. a Call the end points of the diameter A and B. Mark a point (other than A or

B) anywhere on the circumference of the circle and call it C.b Join C to A and B by straight lines.c Mark 2 other points on the circumference. Call them D and E.d Draw angles ADB and AEB and measure them. Compare angles ACB, ADB

and AEB and state your conclusion.

mmultiple choiceultiple choice

2

5

7

m

2

3

7

m 2

3

4

m

21

4 m

A B C D EAll arepossible

5 cm4 cm

9 cmP

Q

ST

R


6

A B

C

DE

O

A B

C

DE

O



History of mathematicsT h e y c o u l d n ’ t d o i t !

The ancient mathematicians solved a great number of problems using just a collapsible compass and an unmarked ruler. They built their whole study of geometry on these tools and the proofs developed using them. There were, however, three problems that they couldn’t solve. They were:1. the trisection of an angle2. the squaring of a circle3. the doubling of a cube.

These problems were studied by the greatest mathematical minds of the time but without success. Their importance in history lies with the fact that they could not be solved and thus they opened up the possibility of other mathematical systems to the mathematicians of the time.

1. The trisection of an angleThis problem involves trying to divide an angle into three equal angles using only a collapsible compass and an unmarked ruler.

Many mathematicians worked on this including Anaxagoras (about 440 BC) while he was in prison. Pappas (about AD300) worked on it and showed how to do it with a marked ruler. The halving of an angle was easily shown using this equipment.

Archimedes showed how to do this with a marked straight-edge in his book of Lemmas.

2. The squaring of a circleThis problem involved creating a square with the same area as a given circle using only a collapsible compass and an unmarked ruler.

Since we know, and they did at the time, that for this we have πr2 = s2, where r is the radius of the circle and s is the side length of a square. Taking the square root of both sides gives us s = r , which means that the ratio between the square side compared to the given (known) radius is proportional to the square root of π.

The Hindu mathematicians worked on this problem from about 800 BC to 500 BC and although they were able to convert rectangles to squares of the same size, they were not able to do it with a circle. Anaxagoras also worked upon

this problem around 440 BC. Archimedes (around 300 BC) worked on a variety of geometrical problems that reduced to this specific problem but he was unable to solve it.

The problem with this example is that there is no way to find the exact square root of π using these tools.

3. The doubling of a cube This involved creating a cube of double the volume of a given cube using only a collapsible compass and an unmarked ruler. This was known as the Delian problem as it was said to have been created by the Oracle at Delphi.

This problem was worked on by Archytas around 400 BC. He worked with the intersection of a cone, a cylinder and a torus but he could not do it with just a compass and ruler.

The problem with this lies in finding the cube root of 2 which cannot be done with these tools.

ProofPierre Wantzel (1814–48) worked on these problems and initially tried to trisect an angle of 60° and found that to do this it was necessary to construct an answer for the equation x3 − 3x − 1 = 0. In 1837 he was able to show that this could not be done and therefore it was not possible to trisect an angle. He then went on to show that the other two problems could not be solved.

Questions1. If you trisected an angle of 60º what size

would each of the three angles be?2. Why was the ‘doubling of a cube’ problem

known as the Delian problem?3. Name a mathematician who tried to solve the

squaring of a circle problem.4. Who showed that these three problems

could not be solved?5. In trying to solve these three problems,

describe what difficulty needed to be overcome in each case.

ResearchUse the Internet or library to find out more about Pierre Wantzel.

π



Angles in a circleIn the diagram at right, chords AC and BC form the angle ACB. Wesay that arc AB has subtended angle ACB.

Theorem 7All angles that have their vertex on the circumference and are subtended by the same arc are equal.

In the diagram at right, angles a, b and c are all subtended by minorarc PQ and, therefore, they are all equal in size; that is, a = b = c.These angles can also be called angles in the same segment.

Theorem 8The angle subtended at the centre of a circle is twice the angle subtended at the circumference, standing on the same arc.

In the diagram at right, angles a and b are both subtended by minorarc PQ with angle a having its vertex on the circumference and angleb having its vertex at the centre. This implies that angle b is twice thesize of angle a, or that b = 2a.

Theorem 9Angles 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 areright angles. This theorem is in fact a special case of the previoustheorem. ∠POQ is a 180° angle at the centre on the arc PQ while a,b and c are all angles at the circumference standing on the same arc.

Theorem 10If 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 cir-cumference. The tangent to the circle is also drawn at P. The radiusand the tangent meet at right angles; that is, the angle at P = 90°.

A B

C

Angles subtended by the same arc

P Q

ab

c

Angles atthe centre and the circumference

P Q

R

a

bO

Angles atthe centreand the circumference — proof

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

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

x = 46° (Angles on the circumference standing on the same arc are equal.)

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° (The angle at the centre is twice the

angle at the circumference standing on the same arc.)

1

2

7WORKEDExample

y

x46°O

OP

c b

a

Q

Tangents

PO



Theorem 11The 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.

In the diagram at right, the line OS bisects ∠RST, formed bythe two tangents. That is, ∠RSO = ∠TSO = a. Moreover, thestraight line OS also bisects ∠ROT, formed by two radii.That is, ∠ROS = ∠SOT = b.

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

THINK WRITEAngle z is subtended by the diameter. Use an appropriate theorem to state the value of z.

z = 90°(Angles in a semicircle are right angles.)

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° (The tangent and the radius drawn to the point of contact meet at right angles.)

1

2

8WORKEDExample

O

zs

S

R

T

O aa

bb

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

THINK WRITEAngles 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° (The angle between the radius and the tangent through the point of contact is a right angle.)

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

∆ABO: t + 90° + 68° = 180° (angle sum of a triangle)t + 158° = 180°

t = 22°∠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 (The angle formed by two tangents is bisected by the line joining its vertex to the centre of the circle.)∠ABO = t = 22°, ∠CBO = u

u = 22°The line BO also bisects ∠AOC, which means that angles q and 68° are equal.

∠AOB = ∠COB (The angle at the centre formed by the radii is bisected by the line from the centre to an external point.)

∠AOB = 68°, ∠COB = qq = 68°

1

2

3

4

9WORKEDExample

B

A

C

O tu

r

q

s

68°



Angles in a circle

1 Find the values of the pronumerals in each of the following, giving reasons for youranswers.

2 Find the values of the pronumerals in each of the following figures, giving reasons foryour answers.

remember1. Angles with their vertices on the circumference, subtended by the same arc, are

equal.2. An angle with its vertex at the centre of the circle is twice the size of an angle

subtended by the same arc, but with the vertex at the circumference.3. Angles subtended by the diameter are right angles.4. A tangent and a radius, drawn to the same point on a circle, meet at a 90°

angle.5. An angle formed by two tangents is bisected by the line joining the vertex of

that angle to the centre of the circle.

remember

10BWWORKEDORKEDEExample

7

x

y40°x

30°

•O

A B

•

x80°

A

B

42°O•x

A

B

x50°

y

O•

x 28°

O•A

B

A B

x30°

yx

P Q

25° x

S

R

32°

g h i

a b c

d e f

SkillSH

EET 10.1


8

x

y

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d e f

• x38°

x

•O

a b c

r

s• ••

u t

•

m n


C h a p t e r 1 0 C i r c l e g e o m e t r y 3733 Given that AB and DB are tangents, find the value of the pronumerals in each of the

following, giving reasons for your answers.

4In the diagram below, which angle is subtended by the same arc as ∠APB?

5Referring to the diagram below, which of the statements is true?

6 Values are suggested for the pronumerals in the diagram below. AB is a tangent to acircle and O is the centre. In each case give reasons to justify suggested values.


9

x

yw

z

O 70°•

A

B

D

t

s

r

O40° •

A

B

D

xyzO 20°•

A

B

D

xs y

zr

O

70°

•

A B

D

y zx

O

20°

•

A B

D

xy

15°

A

D

B

•

Oz

a b

c d

e

f


A

C B

P

D A ∠APCB ∠BPCC ∠ABPD ∠ADBE ∠BAD


A

CB

F

O

D

E

A ∠AED = 2∠ABDB ∠AED = ∠ACDC ∠ABF = ∠ABDD ∠ABD = ∠ACDE ∠ABF = 90°

Ar s

tu

nmC

B

F

O

D

25°

a s = t = 45°b r = 45°c u = 65°d m = 25°e n = 45°


MA

TH

SQUEST

C H A L L

EN

GE

MA

TH

SQUEST

C H A L L

EN

GE


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

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 circleare equal in length.

9 Prove that an angle formed by two tangents is bisected by the line joining the vertexof that angle to the centre of the circle.

10 Use the figure drawn below to prove that angles subtended by the same arc are equal.

P Q

R

a

bO

P QS

R y

O

x

WorkS

HEET10.1

P

SR

O

Q

1 A clock has a minute hand that is 6 cmlong. What is the distance travelled by thetip of the minute hand in one day?

2 A golf ball is packed into a cubical boxso that it just fits. What percentage ofthe box is unfilled?

3 You have six sections of fencing, each3 m long. What is the largest area thatyou can enclose using all six sections?


HoHow mw much of the land on Earuch of the land on Earth was coth was covverered withed with during the last during the last

Hint: Pythagoras’theorem will come in

handy.

The value of the letteredangles and lengths gives the

puzzle answer code.

5 mB m

A

42°

12 m

25 m

58°

K

N m

126 m

H m

16 m

15 m

I m

C38°

T

208°

R

S

150°

78°

E

D

O m

86°

60 m

P

50 m

48 m U

34°

W

Y

105

28 24 51 71 105 51 90 68 28 105 17 63 28 28 43 50 13 48

51 48 39 13 48 71 58 34 48 105 39 15

7m




Cyclic quadrilateralsA 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 circumfer-ence and, 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 12The opposite angles of a cyclic quadrilateral are supplementary (add to 180°).

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

Quadrilaterals in circles1 Construct a circle of radius 10 cm.2 Mark points A, B, C and D on different points on the circumference.3 Join points A and B, B and C, C and D and A and D by straight lines to

construct a quadrilateral.4 Accurately measure the interior angles at points A, B, C and D.5 Repeat steps 1–4 for another circle.6 What is the relationship between

a ∠ABC and ∠ADC?b ∠BAD and ∠BCD?

7 In each circle, extend AD to form an exterior angle at A and measure the exterior angle. What is the relationship between this exterior angle at A and ∠BCD?

A

C

B

D

Cyclic quadrilaterals 1 A

C

B

DCyclic quadrilaterals 2 — proof

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

THINK WRITEPQRS is a cyclic quadrilateral, so its opposite angles are supplementary. First find 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°

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

∠SPQ + ∠QRS = 180°∠SPQ = 120°, ∠QRS = yy + 120° = 180°

y = 60°

1

2

10WORKEDExample

P

R

Q

S

75°120°

x

y


C h a p t e r 1 0 C i r c l e g e o m e t r y 377Theorem 13The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

In the diagram at right, ∠QRS = ∠SPT = a.

Proof: ∠QPS + ∠QRS = 180° (opposite angles of a cyclic quadrilateral)

∠QPS + ∠SPT = 180° (adjacent angles on a straight line).

Therefore ∠SPT = ∠QRS.

Cyclic quadrilaterals

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

P

R

Q

S

b

aaT

Cabri Geometry

Exteriorangles of

cyclicquadrilaterals

Find the value of the pronumerals in the following diagram.

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

x = ∠DAB, ∠DAB = 50° So x = 50°.

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

∠ADC = 100°, ∠ADC = y So y = 100°.

1

2

11WORKEDExample

D B

A

C

y

x100°

50°

remember1. A cyclic quadrilateral has all four of its vertices on the circumference of a

circle.2. Opposite angles of a cyclic quadrilateral are supplementary.3. The exterior angle of a cyclic quadrilateral is equal to the interior opposite

angle.

remember

10CWWORKEDORKEDEExample

10

yx

92°95°

m65°

n

155°

x

50° x yO

x

y

O135°

85°

a b c

d e f



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

3Which of the following correctly states the relationship between x, y and z in the dia-gram shown?

4

a Which of the following statements is always true for the diagram shown?

b Which of the following statements is not correct for the diagram shown?

5 The steps below show you how to set out the proof that the opposite angles of a cyclicquadrilateral are equal.

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

A r = t B r = p C r = q D r = s E r = p + t

A r + p = 180° B q + s = 180° C t + p = 180° D r + s = 180° E t = r


11

x

y80°

85°

x

y

115°

110°x

75°

x150°

x

y 120°

mn

120° 130°

a b c

d e f


z

x

y

O

A x = y and x = 2zB x = 2y and y + z = 180°C z = 2x and y = 2zD x + y = 180° and z = 2xE x + y = 180° and y + z = 180°


s

q r

t p

xA

D

B

C

O

a Find the size of ∠DOB.b Find the size of the reflex angle DOB.c Find the size of ∠BCD.d Find ∠DAB + ∠BCD.

GAME

time

Circle geometry — 001


MA

TH

SQUEST

C H A L L

EN

GE

MA

TH

SQUEST

C H A L L

EN

GE


Find the value of the pronumeral in each of the following.

1

1

4

7

10

2

5

8

3

6

9

a6 cm

1.5 cm2 cm

4 cm5 cm

5 cm

b cm

C

B

A

X

D3 cm

9 cm c

7 cmd

e

28°

f

260°

g

O

h

52°

j

85°

i

79°

1 Each cube in the set of four cubes shown atright is exactly alike. Copy and complete thenet for one of these cubes to show how eachside is lettered.

2 Ten coins are placed as shown inFigure 1. How can you move exactlythree coins to form a triangle like thatshown in Figure 2?

10987

654

32

1Figure 1 Figure 2

AC

F

E

EBB

D

E

A

MQ10 VIC ch 10 Page 379 Tuesday, November 20, 2001 3:34 PM


Great circlesWork that we have done previously on arc lengths enables us to calculate lengths anddistances on the Earth’s surface. This is done by drawing circles on the surface of theEarth and making appropriate calculations.

The Earth can be approximated by a sphere with a radius of about 6400 km. Itrotates about a straight line obtained by joining the north and south poles, as shown.

A great circle is the largest circle that can be formed on the Earth’s surface. Thecentre and radius of a great circle are the same as those of the Earth. The most com-monly known great circle is the equator.

Other great circles pass through the north and south poles.These are called the meridians of longitude.

The prime meridian passes through Greenwich in Londonand is called the Greenwich meridian.

Equator

N

S

Meridian oflongitude

S

N

Gre

enw

ich

mer

idia

n



If a great circle passes through any twopoints on the Earth’s surface, then theshortest distance between the two pointsis the distance measured along the greatcircle.

Longitude and latitudeThe meridians are used to measure longi-tude, while circles parallel to the equatorare used to measure the latitude.

The position of a point on the Earth’ssurface is given by:1. its longitude, which is written as an

angle east or west of the prime mer-idian; and

2. its latitude, which is written as anangle north or south of the equator.

The distance between any two points on agreat circle can be calculated using the for-mula for arc length.

Arc length = × circumference

or l = × 2πr

The radius that needs to be substituted into the formulais the radius of the Earth (6400 km) and the angle isobtained by taking either:

1. the difference between the longitudes of the two pointsif they are on the equator, or

2. the difference between the latitudes of the two points ifthey are on the same meridian.

Greenwich Mean Time1 Locate the Greenwich meridian

(prime meridian) on a map.2 What is Greenwich Mean Time

(GMT)?3 What time is it in Melbourne when it

is 12 noon GMT?4 What line runs from the north pole to

the south pole on the opposite side of the globe to the Greenwich meridian?

5 What is the significance of this line?6 Why is this line not straight?

S

N

Meridianoflongitude

Latitude

sector angle360°

----------------------------

θ360°-----------

l

rθ



Find the distance (to the nearest km) between the following pair of points located on the equator:12°E and 40°W.

THINK WRITEDraw a diagram. Show the prime meridian and the equator. Both points are located on the equator with point 12°E being 12° to the right and point 40°W being 40° to the left of the prime meridian.

Calculate the size of the sector angle, θ. θ = 40° + 12° = 52°

Write the formula for finding the arc length.

l = × 2πr

State the values of θ and r. θ = 52°, r = 6400

Substitute into the formula and evaluate.

l = × 2 × π × 6400

= 5808.456 Write your answer, rounding to the nearest kilometre.

The distance between the two points is 5808 km.

1

2

3θ

360°-----------

4

552°

360°-----------

6

12WORKEDExample

S

O

N

40°12°

Find the distance (to the nearest km) between two points on the same meridian with latitudes 20°N and 36°S.

THINK WRITEDraw a diagram. Show the meridian and the equator. The point 20°N is 20° up and the point 36°S is 36° down the meridian, if measured from the equator.

Calculate the size of the angle subtended by the arc.

θ = 20° + 36° = 56°

Write the formula for the arc length. l = × 2πr

State the values of θ and r. θ = 56°, r = 6400

Substitute the values of θ and r into the formula and evaluate.

l = × 2 × π × 6400

= 6255.260 Write your answer, rounding to the nearest kilometre.

The distance between the points is 6255 km.

1

2

3θ

360°-----------

4

556°

360°-----------

6

13WORKEDExample

S

N

36°

20°O


C h a p t e r 1 0 C i r c l e g e o m e t r y 383It is common to use nautical miles (NM) to express the arclength of a great circle, where 1 NM is the arc length for

a (or 1 minute) angle subtended at the centre of the

Earth. This implies that the arc subtending a 1° angle withthe vertex at the centre of the Earth is 60 NM in length.

On the other hand, using the formula developed previ-ously for the 1° angle and 6400 km radius, the length ofarc is given by:

l = × 2 × π × 6400

= 111.6 km. Therefore 60 NM is approximately equal to 111.6 km.

The length of arc on the great circle, subtending an angle of 1° with the vertex at the centre of the Earth, is equal to 60 NM or 111.6 km.

This knowledge allows us to find the length of arc on the great circle directly (that is,without using the arc length formula) if the angle that the arc subtends at the centre ofthe Earth is given.

160------°

S

O

N

1°

60 NM

1°360°-----------

Find the length of the arc on the great circle subtending the angle of 68° at the centre of the Earth.

THINK WRITEFor an angle of 1°, the arc length is approximately 111.6 km. So to find the length of arc subtending a 68° angle, multiply 68 by 111.6 km.

Arc length = 68 × 111.6 km= 7588.8 km

14WORKEDExample

remember1. The Earth can be approximated by a sphere with a radius of 6400 km.2. An example of a great circle is the equator. 3. Meridians of longitude, passing through the north and south poles, form other

great circles.4. The latitude of a point is its angle to the north or south of the equator.5. The longitude of a point is its angle to the east or west of the Greenwich

meridian (prime meridian).6. The distance between two points on a great circle can be found by using the arc

length formula: l = × 2πr, where θ is the angle subtended by the arc with the

vertex at the centre of the Earth and r is the radius of the Earth (approx. 6400 km).7. The length of arc subtending a 1° angle with the vertex at the centre of the

Earth is approximately 111.6 km or 60 NM.8. The distance (in km) between two points on a great circle can also be found by

using the formula: l = θ × 111.6 where θ is the angle formed at the centre of the Earth.

θ360°-----------

remember



Great circles

1 Find the distance (to the nearest km) between each of the following pairs of pointslocated on the equator. a 70°E and 56°W b 130°E and 160°Wc 46°W and 63°E d 27°W and 142°E

2 Find the distance (to the nearest km) between two points on the same meridian withthe following latitudes.a 15°N and 17°S b 80°N and 20°Sc 50°N and 16°S d 72°N and 26°S

3 Find the length of the arcs on a great circle subtending the following angles at thecentre of the Earth.a 72° b 15° c 12° d 88°

4 Calculate the distance between the following points on the Earth’s surface as shown inthe diagram below.

5

The length of the equator is (use r = 6400 km):

6

All points on the equator have a latitude of:

7 Ravensthorpe (34°S) and Nullagine (22°S) are two towns on the same meridian inWestern Australia. What is the distance between the two towns?

A 10 048 km B 20 096 km C 40 212 km D 12 800 km E Unable to be determined

A 0° B 45° C 90° D 180° E none of those

10DWWORKEDORKEDEExample

12SkillSH

EET 10.2


13


14

Mathca

d

Arc length

S

N

30°

40°

20°

O

P

Q

RT

a P and Tb P and Qc Q and Rd Q and T



GAME

time

Circle geometry — 002


C h a p t e r 1 0 C i r c l e g e o m e t r y 3858 Quito in Ecuador and Macapa in Brazil are on the equator. Macapa has a longitude of

51° W and Quito’s longitude is 78°W. Find the distance between Quito and Macapa.

9 Cardwell and Innisfail are towns 279 km apart in Queensland. They both lie on thesame meridian (146°E). What is the angle between the latitudes of the two towns?

10 The distance between Yarrawonga in Victoria and Griffith in New South Wales is189.8 km. They lie on the same meridian. If Yarrawonga lies on a latitude 36°S, findthe latitude of Griffith.

SOS!

At the beginning of this chapter we considered the search and rescue mission for solo sailor Jesse who has sent a mayday message. Jesse’s coordinates are (26°S, 17°W) and a military vessel heads towards him from coordinates (30°S, 17°W) at a speed of 30 knots. Simultaneously, a search plane takes off from Dakar (14°N, 17°W) at a speed of 750 km/h.1 Calculate the distance of the military vessel from Jesse’s yacht, in nautical

miles.2 Speed on water is usually measured in knots. A speed of 1 knot is equal to 1

nautical mile per hour. Calculate the length of time that it would take for the military vessel to reach Jesse.

3 Calculate the distance of the search plane from Jesse’s yacht, in nautical miles.4 Calculate the distance of the search plane from Jesse’s yacht, in kilometres. Use

this answer to find the length of time that it will take for the plane to reach Jesse.5 Will the military vessel or the plane reach Jesse first? What is the difference in

time taken to reach Jesse?6 Jesse’s yacht will sink in 5 hours, at which time he will need to launch his life

raft. Find:a the speed at which the military vessel would need to travel (in knots) to

reach Jesse in this timeb the speed at which the plane would need to travel to reach Jesse in this time.

WorkS

HEET10.2



Eratosthenes calculates the diameter of the Earth!

In 230 BC, Eratosthenes calculated the size of the Earth. How precise was his measurement?

He knew that during the summer solstice, at noon, the sun shone directly into a well at Syene (now Aswan). At the same time in Alexandria, Egypt, approximately 787 km due north of Syene, he found that the angle of inclination of the sun’s rays was about 7.2°. With these measurements he calculated the diameter and circumference of the Earth.

Since light rays travel parallel to each other, Eratosthenes used pairs of equal

alternate angles and the proportion to find C, the circumference of the

Earth.

1 Explain how Eratosthenes found the angle at the centre of the Earth, subtended by the arc formed from Syene to Alexandria, using the angle of inclination of 7.2°.

2 Explain what the numbers and symbols mean in the proportion . Can you see how Eratosthenes arrived at this?

3 Use the proportion given above to calculate the measurement for the circumference of the Earth that Eratosthenes would have obtained.

4 Use your answer to step 3 to calculate the diameter of the Earth.

Let’s try to calculate the radius of the Earth using Eratosthenes’ measurements but a different method.

5 Use the formula for arc length and the original measurements given to calculate the radius of the Earth (to the nearest km).

6 Hence, calculate the diameter and circumference of the Earth.

7 Compare the results you have obtained so far. What do you notice about the two methods you have used?

8 The Earth’s average radius is usually accepted as 6380 km. Compare this result with that obtained by Eratosthenes. What is the percentage error in Eratosthenes’ result?

r

Centreof Earth

Earth’s surface

Syene

Sun’s rays

Alexandria787 km

7.2°

7.2360--------- 787

C---------=

7.2360--------- 787

C---------=


MA

TH

SQUEST

C H A L L

EN

GE

MA

TH

SQUEST

C H A L L

EN

GE


Find the value of the pronumeral in each of the following, giving a reason for youranswer.

2

1

4

2

5

3

6

In each of the following questions take the radius of the Earth to be 6400 km.

7 Find the distance (to the nearest km) between two points on the equator withlongitude 46°W and 84°W.

8 Find the distance (to the nearest km) between Stockholm (60°N, 18°E) and Budapest(47°N, 18°E)

9 Find the angle subtended at the centre of the Earth between two points on the samemeridian with latitudes 45°S and 13°N.

10 Find the distance between the two points in question 9, correct to the nearest 100kilometres.

k

O

6 cm

n

O

6 cm

4 cm

l

50°

p31°

m

8 cm

q

68°Ο

1 The diameter of the circle shown is 10 cm. What is the area of thesquare ABCD?

2 Imagine a wire tied tightly around the equator of the Earth. a If a second wire also circles the Earth at the equator but is 1 m above

the ground, how much longer is the second wire than the first?b How high above the ground should a wire be placed above the

equator so that it is twice as long as the original wire?

A B

D

O

C



LocusA locus (plural loci) is a set of points that satisfies some condition. It can be consideredas the path created by a moving point. An equation in terms of ordered pairs (or coordi-nates) (x, y) can be used to describe a locus.

Locus of a point equidistant from 2 given points

Find the equation of the locus of P(x, y) given that P is equidistant from R(1, 2) andS(3, −2).

THINK WRITEDraw a diagram to show points S and R and the moving point P.

Since point P is equidistant from points R and S (PR = PS), the distance formula can be used to derive the equation.

Write the formula for the distance between two points.

d =

Identify (x1, y1) and (x2, y2) for S and P.

For S and P: (x1, y1) = (3, −2)(x2, y2) = (x, y)

Substitute the known information into the formula.

dPS =

Repeat steps and for the points R and P.

For R and P: (x1, y1) = (1, 2)(x2, y2) = (x, y)

dPR =

As PS = PR, equate the two expressions.

As PS = PR

=

Simplify by first squaring both sides. =

Expand both sides. x2 − 6x + 9 + y2 + 4y + 4 = x2 − 2x + 1 + y2 − 4y + 4

Move all terms to the left and simplify to get the equation of the locus ofP(x, y).

x2 − 6x + 9 + y2 + 4y + 4 − x2 + 2x − 1 − y2 + 4y − 4 = 0

8y − 4x + 8 = 0 2y − x + 2 = 0(or 2y = x − 2, so y = x − 1)

1

x

y

0

R(1, 2)

S(3, –2)

321–1–2

2

1

–1

–2 P(x, y)

2

3 x2 x1–( )2 y2 y1–( )2+

4

5 x 3–( )2 y 2+( )2+

6 4 5

x 1–( )2 y 2–( )2+

7

x 3–( )2 y 2+( )2+ x 1–( )2 y 2–( )2+

8 x 3–( )2 y 2+( )2+ x 1–( )2 y 2–( )2+

9

10

12---

15WORKEDExample

Equation of the locus of a point equidistant from 2 given points



A set of points equidistant from a given point (the centre) represents a circle. The equation of a circle with centre (0, 0) and radius, r, is:

x2 + y2 = r2.

The equation of a circle with centre (h, k) and radius, r, is:

(x − h)2 + (y − k)2 = r2.

Find the equation of the locus of a point P(x, y) such that P is 5 units from the y-axis.

THINK WRITEDraw a diagram. The locus of point P is represented by 2 straight lines passing through 5 and −5 on the x-axis. Any point on either of the two lines is 5 units from the y-axis.

The equation of the vertical line crossing the x-axis at a is x = a. So state the equations of the two lines in question.

The equations are:x = 5 and x = −5.

1

x

y

5–5

2

16WORKEDExample

Cabri Geometry

Locus of apoint

equidistantfrom a

given point

Find the equation of each of the following loci:a a set of points 5 units away from (0, 0)b a set of points 3 units away from the point (3, 2).

THINK WRITEa The required set of points is a circle

of radius 5 units and centre (0, 0). Write the equation of a circle.

a x2 + y2 = r2

State the value of r. r = 5

Substitute 5 for r into the equation and simplify.

x2 + y2 = 52 x2 + y2 = 25

b The required set of points is a circle of radius 3 units and centre (3, 2). Write the equation of a circle with the centre at (h, k).

b (x − h)2 + (y − k)2 = r2

State the values of h, k and r. h = 3, k = 2, r = 3

Substitute the values of h, k and r into the equation of a circle and simplify. (There is no need to expand the brackets.)

(x − 3)2 + (y − 2)2 = 32

(x − 3)2 + (y − 2)2 = 9

1

2

3

1

2

3

17WORKEDExample



The circle equation is more difficult to enter into a graphics calculator than most equa-tions because it is not in the form y = . . . We can, however, graph a circle using thecalculator’s DRAW function.

Suppose that we want to draw the graph of x2 + y2 = 16. This is a circle with itscentre at (0, 0) and a radius of 4 units.1. Press [DRAW] and select 9:Circle(:.2. Enter the coordinates of the centre and the radius, separated by commas.

3. Press and the circle will be drawn. It may be, however, that the circleappears elliptical rather than perfectly round. This is because of the scale used by thegraphics calculator.

This can be fixed using the ZOOM function. 4. Press and select 5: Zsquare. This will readjust the display with an equal

scale on each axis. Draw the circle again. The circle should now appear round.

5. Use the DRAW function to graph the circles:(a) x2 + y2 = 36(b) (x − 5)2 + (y − 2)2 = 9(c) x2 + (y + 2)2 = 49

Graphics CalculatorGraphics Calculator tip!tip! Graphingcircles

2nd

ENTER

ZOOM



Locus

1 Find the equation of the locus of point P(x, y) given that P is equidistant from R and S,where:a R is (3, 4) and S is (4, −5)b R is (−1, 2) and S is (3, 1)c R is (−2, −3) and S is (2, 3).

2 Find the equation of the locus of a point P(x, y) such that:a P is 4 units from the y-axisb P is 2 units from the x-axisc P is 10 units from the line x = 1.

3 Find the equation of each of the following loci.a A set of points 3 units away from (0, 0)b A set of points 7 units away from (0, 0)c A set of points 1 unit away from (0, 0)

4 Find the equation of each of the following loci.a A set of points 3 units away from the point (2, 1)b A set of points 6 units away from the point (−3, −2)c A set of points 4 units away from the point (3, −1)

5The centre of a circle with the equation x2 + (y + 2)2 = 14 is at:

6The radius of the circle with the equation given in question 5 is:

7 A point moves so that its perpendicular distance from the x-axis is equal to its distancefrom point (2, 1). Find the equation of this locus.

8 Find the equations of the circles with the following centres and radii:a centre (0, 0), radius 6 b centre (−2, −1), radius 4 c centre ( , ), radius .

A (0, 0) B (−2, 0) C (0, 2) D (0, −2) E (2, 0)

A 0 B 14 C 2 D E 7

remember1. A locus is a set of points (a path of a moving point) which satisfies some

condition.2. The distance between two points A(x1, y1) and B(x2, y2) is given by

.3. The equation of a circle with the centre at (0, 0) and radius, r, is given by:

x2 + y2 = r2.4. The equation of a circle with radius, r, and centre at (h, k) is given by:

(x − h)2 + (y − k)2 = r2.

d x2 x1–( ) y2 y1–( )+=

remember

10EWWORKEDORKEDEExample

15

SkillSH

EET 10.3

SkillSH

EET 10.4


16

Cabri Geometry

Equation ofthe locus of

a pointequidistant

from 2 givenpoints


17a


17b

Cabri Geometry

Locus of apoint

equidistantfrom a given

point



Mathcad

Locus

14

WorkS

HEET10.3

12--- 1

3--- 5



Going loci 1 The photograph below shows hydraulic shovels. Consider a point on the bottom

‘lip’ of a shovel. Describe the path traced out (locus) by this point while the machine is working.

2 This photograph shows a bicycle wheel. Consider a point on the edge of the wheel. Describe the locus of the point as it moves along the road.

3 Find some other examples of moving objects that trace out a recognisable locus as they move.



Copy the sentences below. Fill in the gaps by choosing the correct word or expression from the word list that follows.

1 The outside length or boundary forming a circle is called the .2 A straight line from one point on the circumference to another is a .3 A is an area formed by a chord and part of the circumference.4 An area of a circle enclosed by two radii and part of the circumference is

called a .5 A is formed when a chord is extended beyond the circumfer-

ence on one side.6 A radius meeting a chord at 90° the chord.7 A and a radius drawn to the same point on the circumfer-

ence of a circle meet at a 90° angle.8 Angles with a vertex on the circumference, subtended by the same arc,

are .9 An angle with the vertex at the centre is the angle subtended

by the same arc but with the vertex at the circumference.10 Angles subtended by the diameter are .11 Opposite angles of a cyclic quadrilateral are .12 The angle of a cyclic quadrilateral is equal to the interior

opposite angle.13 The largest circles formed on the Earth’s surface are called .14 An angle east or west of the prime meridian represents the .15 An angle north or south of the equator represents the .16 Distances on the Earth’s surface can be calculated using the formula for

: l = × 2πr where θ is the angle subtended by the arc with

the vertex at the centre of the Earth and r is the radius of the Earth.17 The radius of the Earth is approximately .18 A is a set of points satisfying some condition.19 The equation of a circle with centre (0, 0) and radius, r, is given by .20 A circle with centre at (h, k), and radius, r, has the equation .21 The distance between two points A(x1, y1) and B(x2, y2) is given by .

summary

q360∞-----------

W O R D L I S Ttangentsecantlocus6400 kmsupplementaryexteriorequal in size

longituderight angles circumferencesegmentchord

twice

sector great circlesx2 + y2 = r2 (x − h)2 + (y − k)2 = r2

bisectslatitude arc length

x2 x1–( ) y2 y1–( )+



1 Find the value of m in each of the following.

2 Find the value of the pronumerals in the following.

10A

CHAPTERreview

10

6

5

m

4

8

6

m

4

5

3m

812

10m

a b

c d

10A

4

6

x

4

8

k

4

5

mn

7

x

1

5.5

811

6 b

a

3

2

w

x

a b c

d

ef



In which of the following figures is it impossible to get a reasonable value for the pronumeral?

4

Which of the following statements is NOT true for the diagram shown?

5 Two chords, AB and CD, intersect at E as shown. If AE = CE, prove that EB = ED.

6 Two circles intersect at X and Y. Two lines, AXB and CXD, intersect one circle at A and C, and the other at B and D, as shown. Prove that ∠AYC = ∠BYD.

10Ammultiple choiceultiple choice

56

m

4

2

6

m

3

28

m

5

4

7m

3

1

7m

2

A B C

D E

10Ammultiple choiceultiple choice

B

CO

AA AO = BOB AC = BCC ∠OAC = ∠OBCD ∠AOC = 90°E ∠OBC = 90°

10A

A

B

C

E

D

10A

A

BC

D

X

Y



7 Determine the values of the pronumerals in each of the following.

8 Find the value of the pronumeral in each case.

9 Name at least five pairs of equal angles in the following diagram.

10Ba b c

d e f

g h i

j k l

x50°

xy

25° 48°x

z

y28°O

x

O

y

O xO

35°

x

O

110°

x

O

250°

x110°

x

100°

m 70°

y

x

z40°30°

O

10Bx

O xO 70°

x O110°

x O70°

a b

c d

10BC

U

O

P

Q

R

S

T


C h a p t e r 1 0 C i r c l e g e o m e t r y 39710 Find the values of the pronumerals in the following figures.

11

Which of the following statements is NOT always true for the diagram below?

12 An arc on a great circle subtends an angle of 72° at the centre of the Earth. Find the arc’s length.

13 Find the distance (to the nearest km) between the following pairs of points on the equator:a 50°E and 25°W b 125°E and 36°W.

14 Find the distance (to the nearest km) between the following pairs of points, given the latitudes on the same meridian:a 17°N and 28°S b 80°N and 27°S.

10C

y

x

100°

85° x 81°

y

x

78°

92° yx

88°

97°

a b

c d

10Cmmultiple choiceultiple choice

c

ba

d

e

A ∠a + ∠c = 180°B ∠b + ∠d = 180°C ∠e + ∠c = 180°D ∠a + ∠e = 180°E ∠a = ∠e

10D

10D

10D

MQ10 VIC ch 10 7 Tuesday, November 20, 2001 3:34 PM


15 Calculate the distance between these points on the Earth’s surface.a points M and Nb points N and R

16 Thargomindah and Yaraka are towns in Queensland. Both lie on the 144°E longitude. Thargomindah lies on a latitude of 28°S and Yaraka is on a latitude of 25°S. Calculate the distance between the two towns.

17 Find the equation of the locus of point P(x, y), given that P is equidistant from R(1, 5) and S(2, −4).

18 Find the equation of the locus for a set of points 5 units away from (0, 0).

19 Find the equations of the following circles.a centre (0, 0), radius 6b centre (1, −2), radius 3

20

The centre of a circle with the equation (x + 1)2 + (y − 3)2 = 20 is at:A (1, −3) B (0, 0) C (−1, −3) D (−1, 3) E (1, 3)

10D

South

O

North

28°

10°

MNR

10D

10E

10E10E

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10

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