Loading...

Properties of the Circle TERMINOLOGY Arc: Part of a curve, most commonly a portion of the distance around the circumference of a circle Chord: A straight line joining two points on the circumference of a circle Concentric circles: Circles that have the same centre Concyclic points: Points that lie on the circumference of the same circle Cyclic quadrilateral: A cyclic quadrilateral is a figure whose four vertices are concyclic points. The four vertices lie on the circumference of a circle

Radius: A radius is the distance from the centre of a circle out to the circumference (radii is plural, meaning more than one radius) Subtend: Form an angle at some point (usually the centre or circumference of a circle) Tangent: A straight line external to a curve or circle that just touches the curve or circle at a single point

Chapter 9 Properties of the Circle

INTRODUCTION IN CHAPTER 4, YOU STUDIED the geometry of angles, triangles, quadrilaterals

and other polygons. This chapter shows you some properties of the circle.

DID YOU KNOW? A rainbow is the shape of an arc of a circle. If you could see the whole rainbow, it would form a circle. Research the rainbow on the Internet and find out more about its shape and other properties.

Parts of a Circle An arc is a part of the circumference.

A tangent touches the circle at one point.

499

500

Maths In Focus Mathematics Extension 1 Preliminary Course

Concentric circles are circles that have the same centre.

Equal circles have the same radius.

DID YOU KNOW? Leonardo da Vinci (1452–1519) was a great artist, scientist and inventor. He studied geometry, and many of his model drawings show this influence. His drawings included designs for flying machines, spring-driven automobiles, bridges and weapons. Leonardo’s designs were revolutionary, and the scientists of his time did not have the knowledge needed to make the models work.

Arcs, Angles and Chords Equal arcs subtend equal angles at the centre of the circle.

Chapter 9 Properties of the Circle

501

Proof Let two equal arcs have lengths l 1 and l 2, and subtend angles of a and b at the centre of the circle. Using l = ri , l 1 = ra and l 2 = rb But l 1 = l 2 ` ra = rb ` a=b The converse is also true:

If two arcs subtend equal angles at the centre of the circle, then the arcs are equal.

EXAMPLE

AB is a diameter of the circle with centre O. Arc CB = arc BD. Prove +AOC = +AOD.

Solution Since arc CB = arc BD,+COB = +DOB Let +COB = +DOB = x Then +AOC = 180c - +COB = 180c - x Also +AOD = 180c - +DOB = 180c - x ` +AOC = +AOD

(+AOB is a straight angle) (similarly)

You will study the formula l = r i in Chapter 5 of the HSC Course book.

502

Maths In Focus Mathematics Extension 1 Preliminary Course

Equal chords subtend equal angles at the centre of the circle.

Proof OA = OC OB = OD AB = CD ` by SSS, DOAB / DOCD ` +AOB = +COD

(equal radii) (similarly) (given) (corresponding +s in congruent Ds)

The converse in also true:

Equal angles subtended at the centre of the circle cut off equal chords.

EXAMPLE

AD is a diameter of the circle with centre O, where AB = CD. Prove that BC < AD.

Chapter 9 Properties of the Circle

503

Solution Since AB = CD, +AOB = +COD Let +AOB = +COD = x Then +BOC = 180c - (x + x) = 180c - 2x OB = OC ` D OBC is isosceles with +OBC = +OCB +OBC + +OCB + 180c - 2x = 180c +OBC + +OCB = 2x ` +OBC = +OCB = x ` +OBC = +AOB

(+AOD is a straight +) (equal radii) (+ sum of DOBC)

But these are equal alternate angles ` BC < AD

The angle at the centre of a circle is twice the angle at the circumference subtended by the same arc.

These figures show that this property can look quite different in different situations.

Proof

Join AO and produce to D. Let +BAO = x and +CAO = y ` +BAC = x + y OA = OB

(equal radii)

504

Maths In Focus Mathematics Extension 1 Preliminary Course

` +OBA = x = +BAO OA = OC

(base +s of isosceles D)

` +OCA = y = +CAO

(base +s of isosceles D)

(equal radii)

+BOD = x + x = 2x

(exterior + of DOBA)

+COD = y + y = 2y +BOC = +BOD + +COD = 2x + 2y = 2 (x + y ) = 2+BAC

(exterior + of DOCA)

EXAMPLES 1. Find the values of x and y.

Solution x = 40 (+ at centre is twice the + at the circumference) Reflex +BOC = 360c - 80c (+ of revolution) = 280c ` y = 140 (+ at centre is twice the + at the circumference) 2. Prove +BOC is twice the size of +OCA.

Chapter 9 Properties of the Circle

Solution Let +OAC = x +BOC = 2+OAC = 2x ` `

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

OA = OC ^ equal radii h +OCA = +OAC = x (base +s of isosceles D) +BOC = 2+OCA

Angles in the same segment of a circle are equal.

Proof

Join A and D to centre O +AOD = 2+ABD +AOD = 2+ACD ` +ABD = +ACD

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

505

506

Maths In Focus Mathematics Extension 1 Preliminary Course

EXAMPLE Prove D ABC and D DEC are similar. Hence ﬁnd the value of y correct to 1 decimal place.

Solution +ABC = +DEC +BCA = +ECD

(+s in same segment) (vertically opposite +s)

` D ABC ||| D DEC EC ED ` = BA BC y 6 .5 = 1 .7 1.2 1 .2y = 1 . 7 # 6. 5 1 .7 # 6 .5 y= 1 .2 = 9.2 cm correct to 1 decimal place.

The angle in a semicircle is a right angle.

Proof

Chapter 9 Properties of the Circle

+AOB = 180c +AOB = 2+ACB ` +ACB = 90c

(straight +) (+ at centre is twice the + at the circumference)

EXAMPLE AB is a diameter of the circle below. If BC = 2 m and AB = 9 m, ﬁnd the exact length of AC.

Solution +ACB = 90c ` AB 2 = AC 2 + BC 2 9 2 = AC 2 + 2 2 81 = AC 2 + 4 77 = AC 2 ` AC =

77 m

9.1 Exercises 1.

Find values of all pronumerals (O is the centre of each circle). (a)

(b)

507

508

Maths In Focus Mathematics Extension 1 Preliminary Course

(c)

3.

Find values of all pronumerals (O is the centre of each circle). (a)

(d)

(b)

(e) (c)

(f) (d)

2.

The circumference of a circle is 16r cm. Find the length of the arc that subtends an angle of 40c at the centre of the circle.

(e)

Chapter 9 Properties of the Circle

(f)

4.

Find values of all pronumerals (O is the centre of each circle). (a)

(g)

(b)

(h)

(c)

(i)

(d)

(j)

509

510

Maths In Focus Mathematics Extension 1 Preliminary Course

(e)

(f)

(j)

5.

(a) Prove D ABC ||| D DEC. (b) Hence ﬁnd the value of x correct to 1 decimal place.

6.

Find x and y, giving reasons.

7.

Find x and y, giving reasons.

(g)

(h)

(i)

Chapter 9 Properties of the Circle

8.

Evaluate x, giving reasons for each step in your calculation.

11. Find x, giving reasons for each step in your calculations.

9.

Prove D STV and DWUV are similar. Hence ﬁnd x.

12. The circle below has centre O. D

C

O y A

x 38c B

10. AB = 6 cm and BC = 3 cm. O is the centre of the circle. Show that the radius of the circle is 3 5 cm. 2

(a) Evaluate x and y. (b) Show that AD = BC. 13. Show that AD < BC in the circle below. A

D

33c

114c

C

B

511

512

Maths In Focus Mathematics Extension 1 Preliminary Course

14. A circle has centre O and radius r as shown.

15. The circle below has centre O and +DAB = i. A i

A

B

O

C

B

(a) Show that triangles AOB and ABC are similar. (b) Show that BC = 2 r.

O

C

Show that +DAB and +BCD are supplementary.

Chord Properties A perpendicular line from the centre of a circle to a chord bisects the chord.

Proof

D

Chapter 9 Properties of the Circle

+ADO = +BDO = 90c OA = OB

(given) (equal radii)

OD is common ` by RHS DOAD / DOBD ` AD = BD

(corresponding sides in congruent Ds)

So OD bisects AB The converse is also true: A line from the centre of a circle that bisects a chord is perpendicular to the chord.

EXAMPLES 1. Line OC is perpendicular to chord AB. If the radius of the circle is 6 cm and the chord is 11 cm long, ﬁnd the length of OC, correct to 1 decimal place.

Solution AB = 11 ` AC = 5.5 Also OA = 6

(OC bisects AB) (radius-given)

OA = AC + OC 6 2 = 5.5 2 + OC 2 36 = 30.25 + OC 2 5.75 = OC 2 2

`

2

2

OC = 5.75 = 2.4 cm CONTINUED

513

514

Maths In Focus Mathematics Extension 1 Preliminary Course

2. Given XY = YZ and +OXY = 54c, ﬁnd +XOY.

Solution +OYX = 90c

(OY bisects XZ)

+XOY = 180c - (90c + 54c) = 36c

(+ sum of DOXY)

Equal chords are equidistant from the centre of the circle.

Proof

Let CD = AB +OEB = +OFD = 90c OB = OD AB = CD

(given) (equal radii) (given)

Chapter 9 Properties of the Circle

1 AB 2 1 DF = CD 2 BE = DF BE =

`

by RHS DOEB / DOFD OE = OF

` `

(OE bisects AB) (OF bisects CD)

(corresponding sides in congruent Ds)

The converse is also true:

Chords that are equidistant from the centre are equal.

Class Exercise Prove that chords that are equidistant from the centre are equal.

EXAMPLE In the circle below, with centre O, OE = 35 mm, DE = 56 mm and OC = 21 mm. Show that AB = DE.

Solution EF = 28 mm

(OF bisects DE)

OE = EF + OF 35 2 = 28 2 + OF 2 35 2 - 28 2 = OF 2 441 = OF 2 2

` ` So

2

OF = 441 = 21 mm OF = OC AB = DE.

2

(chords equal when equidistant from the centre)

515

516

Maths In Focus Mathematics Extension 1 Preliminary Course

The products of intercepts of intersecting chords are equal.

AE: EB = DE: EC

Proof

+AED = +CEB +DAE = +ECB +ADE = +EBC

(vertically opposite +s) (+s in same segment) (similarly)

D AED <;D CEB AE DE ` = EB EC ` AE: EB = DE: EC `

EXAMPLE

Given chord AB = 12.3 cm, EB = 2.7 cm and DE = 10.6 cm, ﬁnd the length of EC, correct to 1 decimal place.

Chapter 9 Properties of the Circle

Solution AE = AB - EB = 12.3 - 2.7 = 9 .6 AE: EB = DE: EC 9.6 # 2.7 = 10.6 # EC 9 .6 # 2 .7 ` EC = 10.6 = 2.4 cm

9.2 Exercises 1.

Find the values of all pronumerals (O is the centre of each circle).

(d)

(a)

(e)

(b)

(f)

(c)

517

518

Maths In Focus Mathematics Extension 1 Preliminary Course

(g)

3.

A circle with radius 89 mm has a chord drawn 52 mm from the centre. How long, to the nearest millimetre, is the chord?

4.

O and P are the centres of intersecting circles with radii 20 cm and 8 cm respectively. If AB = 10 cm, find the distance OP, correct to 1 decimal place.

5.

Show AB = CD.

6.

AC = 20 cm and AD = 26 cm. Find OB, correct to 1 decimal place.

(h)

(i)

(j)

2.

Find the exact radius of a circle with a chord that is 8 cm long and 5 cm from the centre.

Chapter 9 Properties of the Circle

7.

Evaluate x and y, correct to 1 decimal place.

9.

A circle with centre O has radius r and chord AB = x. D

O

A

C

8.

B

Find the values of all pronumerals.

Show that CD =

2r + 4r 2 - x 2 . 2 A

10.

E C D

B

(a) Prove that triangles ABC and CDE are similar. (b) Show that AC : CD = BC : CE.

Concyclic Points Concyclic points are points that lie on the circumference of a circle.

519

520

Maths In Focus Mathematics Extension 1 Preliminary Course

Any 3 non-collinear points are concyclic. They lie on a unique circle, with centre at the point of intersection of the perpendicular bisectors of the intervals joining these points.

Four or more non-collinear points may not necessarily lie on a circle.

Cyclic quadrilaterals A cyclic quadrilateral is a ﬁgure whose 4 vertices are concyclic points.

The opposite angles in a cyclic quadrilateral are supplementary.

Chapter 9 Properties of the Circle

Proof

Join B and D to O. Obtuse +DOB = 2+A Reflex +DOB = 2+C Obtuse +DOB + reflex +DOB = 360c ` 2+A + 2+C = 360c ` +A + +C = 180c

(+ at centre is doubl e +at circumference) (+ of revolution)

Similarly, it can be proven that +B + +D = 180c by joining A and C to O. The converse is also true:

If the opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic.

The property of opposite angles being supplementary in a cyclic quadrilateral can also be used to prove the following property:

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

521

522

Maths In Focus Mathematics Extension 1 Preliminary Course

Proof

Let Then `

+A = x +BCD = 180c - x

(opposite +s supplementary in cyclic quadrilateral) (+DCE straight angle)

+BCD + +BCE = 180c

`

+BCE = 180c - (180c - x) = 180c - 180c + x =x +A = +BCE

`

EXAMPLE Evaluate a, b and c.

Solution OB = OD ` +ODB = 20c a + 20c + 20c = 180c ` a = 180c - 40c = 140c b = 70c c = 180c - 70c = 110c

(equal radii) (base + s of isosceles D equal) (+ sum of D)

(+ at centre double + at circumference) (opposite + s in cyclic quadrilateral)

Chapter 9 Properties of the Circle

9.3 Exercises 1.

Find the values of all pronumerals.

(e)

(a)

(f) (b)

(g) (c)

(h)

(d)

(i)

523

524

Maths In Focus Mathematics Extension 1 Preliminary Course

(e)

(j) 118c

yc xc

O 46c

2.

(f) Find the values of all pronumerals. (a)

(g)

(b)

(h)

(c)

(i)

(d)

(j)

xc 111c

Chapter 9 Properties of the Circle

3.

Show that ABCD is a cyclic quadrilateral. (a)

A

i

C

i

58c C C

B

B

D

B 58c

(b)

(c)

525

D

A E

D

A

Tangent Properties

The tangent to a circle is perpendicular to the radius drawn from the point of contact.

The perpendicular distance is the shortest distance — any other distance would be greater than the radius.

The converse is also true: The line perpendicular to the radius at the point where it meets the circle is a tangent to the circle at that point.

526

Maths In Focus Mathematics Extension 1 Preliminary Course

Here is another property of tangents to a circle: Tangents to a circle from an exterior point are equal.

Proof

Join OB +A = +C = 90c OB is common OA = OC ` by RHS, DOAB / DOCB ` AB = CB

(tangent = radius) (equal radii) (corresponding sides in congruent Ds)

EXAMPLE A circle with centre O and radius 10 cm has a tangent AB drawn to it where OB = 26 cm. Find the length of AB.

Chapter 9 Properties of the Circle

527

Solution OA = 10 +OAB = 90c

(radius) (tangent = radius)

` OB 2 = OA 2 + AB 2 26 2 = 10 2 + AB 2 676 = 100 + AB 2 576 = AB 2 ` AB = 576 = 24 cm

When two circles touch, the line through their centres passes through their point of contact.

Proof

You could also prove this result for when circles touch internally.

AB is a tangent to circle with centre O ` +OCB = 90c AB is a tangent to circle with centre P ` +PCB = 90c +OCB + +PCB = 90c + 90c = 180c ` OCP is a straight line.

(tangent = radius) (similarly)

528

Maths In Focus Mathematics Extension 1 Preliminary Course

EXAMPLE Two circles touch at A and the larger circle has centre O. Prove (a) TABC and TAOD are similar (b) CB < DO (c) BC = 2DO.

Solution (a) OA is a diameter of small circle

(line through centres passes through point of contact) (+ in semicircle)

` +ODA = 90c Since AB is a diameter of the larger circle, +BCA = 90c (similarly) ` +BCA = +ODA +A is common ` D ABC ;;; D AOD (b) +BCA = +ODA These are equal corresponding angles. ` CB < DO (c)

AB AB ` OA AB ` OA ` BC

= 2 OA

[from (a)]

(OA radius)

=2 AC BC = =2 AD DO = 2DO =

(by similar Ds)

The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment.

Chapter 9 Properties of the Circle

Proof

Draw in diameter CF and join EF. Let +ECB = x. +FCB = 90c ` +FCE = 90c - x +FEC = 90c

(tangent = radius) (+ in semicircle)

` +EFC = 180c - (90c + 90c - x) (angle sum of D ) =x +EFC = +EDC (angles in same segment) ` +EDC = +ECB

EXAMPLE Evaluate x and y.

CONTINUED

529

530

Maths In Focus Mathematics Extension 1 Preliminary Course

Solution +ABC = 180c - (95c + 50c) = 35c ` x = 35c y = 35c

(+ sum of D ) (+ s in alternate segment) ( y and +ABC in same segment)

The square of the length of the tangent from an external point is equal to the product of the intercepts of the secant passing through this point.

PQ 2 = QR:QS where PQ is a tangent to the circle.

Proof

The third pair of angles is equal by angle sum of a triangle.

+QPR = +PSR +Q is common ` PQR ;;;D SPQ PQ QR PR = = ` QS SP PQ PQ QR ` = QS PQ 2 PQ = QR:QS

(angles in alternate segments)

Chapter 9 Properties of the Circle

EXAMPLE AB is a tangent to the circle and CD = 1.3 cm, BC = 1.7 cm. Find the length of AB, correct to 1 decimal place.

Solution BD = 1.3 + 1.7 =3 AB 2 = BC: BD = 1 .7 # 3 = 5 .1 AB = 5.1 = 2.3 cm correct to 1 decimal place.

9.4 Exercises 1.

Find the values of all pronumerals. (a)

(b)

531

532

Maths In Focus Mathematics Extension 1 Preliminary Course

(c)

(h)

(d)

(i)

y x

38c O

(e) (j)

(f)

2. (g)

Find the values of all pronumerals (all external lines are tangents to the circles). (a)

Chapter 9 Properties of the Circle

(g)

(b)

(h) (c)

(i) (d)

xc

63c

O

yc

(j)

9 cm

15

cm

13 cm

(e)

O

y P

(f)

533

534

Maths In Focus Mathematics Extension 1 Preliminary Course

3.

4.

Find the values of all pronumerals, giving reasons for each step of your working (O is the centre of circle, AB is a tangent).

AB, BC and AC are tangents, with AB = 24 cm, BC = 27 cm and BM = 15 cm. Find the length of AC.

6.

Find the values of all pronumerals (O is the centre of each circle; all external lines are tangents). (a)

(b)

(c)

5.

AB = 6.5 m, AC = 3.9 m and BC = 5.2 m. Prove A lies on a diameter of the circle, given BC is a tangent to the circle. (d)

Chapter 9 Properties of the Circle

(e)

(i)

(f)

(j)

(g)

7.

Find the values of all pronumerals. (a)

(h)

(b)

535

536

Maths In Focus Mathematics Extension 1 Preliminary Course

(c)

(h)

(d)

(i)

(e) (j)

(f) 8.

(g)

Find AB, given AD = 4.9 m, BC = 5.1 m and CD = 7.8 m.

Chapter 9 Properties of the Circle

Test Yourself 9 1.

O is the centre of the circle. Evaluate i.

2.

Evaluate y to 1 decimal place.

3.

4.

5.

Evaluate x, y and z, giving reasons for each step of your working.

6.

O is the centre of the larger circle. Find the value of x.

7.

AB is a tangent to the circle. Evaluate a, b and c.

8.

O is the centre of the circle, and AB is a tangent. Evaluate a, b, c and d, giving reasons for each step of your working.

AB is a tangent to the circle. Find the value of x to 1 decimal place.

O is the centre of the circle. Find the length of tangents x and y.

537

538

Maths In Focus Mathematics Extension 1 Preliminary Course

9.

Find the length of the radius of the circle. O is the centre.

13. Calculate the length of AB to 3 significant figures, given that A and B are the centres of the circles.

10. Find the length of tangent AB. 14. Find the value of x to 1 decimal place.

11. Evaluate a and b.

12. O is the centre of the circle. Find the value of a and b.

15. Find the length of AB.

16. Evaluate a and b.

Chapter 9 Properties of the Circle

17. Evaluate x and y, giving reasons for your working.

18. Evaluate x, y and z.

19. Prove that DBCD is similar to DABC.

20. O is the centre of the circle. (a) Prove that DOAC and DOBC are congruent. (b) Show that OC bisects AB.

Challenge Exercise 9 1.

Find the length of the radius, to the nearest centimetre, if AC = 10 cm and BD = 3 cm.

2.

In the circle below with centre O, OD = DC. Prove +AOE = 3+DCB.

539

540

Maths In Focus Mathematics Extension 1 Preliminary Course

3.

Two circles meet at points A and B. A tangent to each circle is drawn from A to meet the circles at D and C. Prove +DAC = 90c.

5.

The triangles below have +BDE = 2+BAD and +CDE = 2+CAD. Prove that a circle can be drawn through A, B and C with centre D.

4.

Three equal circles touch each other, as in the figure. (a) Prove that the triangle with vertices the centres of the circles is equilateral. (b) Find the perimeter of the curved figure DEF in terms of the radius r of the circles. (c) Find the exact area of the shaded region.

6.

Two chords AB and CD intersect at 90c Prove, for obtuse +AOD,+AOD + +COB = 180c where O is the centre of the circle.

7.

Prove that any kite ABCD with +ADC = +ABC = 90c is a cyclic quadrilateral with diameter AC.

Chapter 9 Properties of the Circle

8.

A large circle with radius R is surrounded by 7 smaller circles with radius r. A circle is drawn through the centres of the smaller circles. If R = area in terms of r.

3r , find the shaded 2

9.

Prove that if an interval subtends equal angles at two points on the same side of it, then the endpoints of the interval and the two points are concyclic.

10. Prove that if both pairs of opposite angles in a quadrilateral are supplementary, then the quadrilateral is cyclic.

541

9
Properties of the Circle TERMINOLOGY Arc: Part of a curve, most commonly a portion of the distance around the circumference of a circle Chord: A st...