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CHAPTER 28 THE CIRCLE AND ITS PROPERTIES

EXERCISE 118 Page 277

1. Calculate the length of the circumference of a circle of radius 7.2 cm. Circumference, c = 2πr = 2π(7.2) = 45.24 cm 2. If the diameter of a circle is 82.6 mm, calculate the circumference of the circle. Circumference, c = 2πr = πd = π(82.6) = 259.5 mm. 3. Determine the radius of a circle whose circumference is 16.52 cm.

Circumference, c = 2πr from which, radius, r = 16.522 2cπ π= = 2.629 cm

4. Find the diameter of a circle whose perimeter is 149.8 cm. If perimeter, or circumference, c = πd, then 149.8 = πd

and diameter, d = 149.8π

= 47.68 cm

5. A crank mechanism is shown below, where XY is a tangent to the circle at point X. If the

circle radius 0X is 10 cm and length 0Y is 40 cm, determine the length of the connecting rod

XY.

If XY is a tangent to the circle, then ∠0XY = 90°

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Thus, by Pythagoras, 2 2 20 0Y X XY= + from which, XY = ( )2 2 2 20 0 40 10 1500Y X− = − = = 38.73 cm 6. If the circumference of the Earth is 40 000 km at the equator, calculate its diameter.

Circumference, c = 2πr = πd from which, diameter, d = 40 000cπ π= = 12 732 km = 12 730 km,

correct to 4 significant figures 7. Calculate the length of wire in the paper clip shown below. The dimensions are in millimetres.

Length of wire = (12 – 2.5) + ( )1 2 2.52

π × + (32 – 2.5 – 6 – 2.5) + ( )1 2 2.52

π × + (32 – 2.5 – 3)

+ ( )1 2 32

π × + 3 + 12

= 9.5 + 2.5π + 21 + 2.5π + 26.5 + 3π + 3 + 12 = 72 + 8π = 97.13 mm

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EXERCISE 119 Page 278

1. Convert to radians in terms of π: (a) 30° (b) 75° (c) 225°

(a) 30° = 30180π

× rad = 6π rad

(b) 75° = 75180π

× rad = 512π rad

(c) 225° = 225180π

× rad = 45 15rad rad36 12π π

= = 54π rad

2. Convert to radians, correct to 3 decimal places: (a) 48° (b) 84°51' (c) 232°15'

(a) 48° = 48180π

× rad = 0.838 rad

(b) 84°51′ = 5184 84.8560 180 180

π π × = ×

rad = 1.481 rad

(c) 232°15′ = 232.25180π

× rad = 4.054 rad

3. Convert to degrees: (a) 76π rad (b) 4

9π rad (c) 7

12π rad

(a) 7 rad6π = 7 180

6π

π°

× = 7 × 30 = 210°

(b) 4 rad9π = 4 180

9π

π°

× = 4 × 20 = 80°

(c) 7 rad12π = 7 180

12π

π°

× = 7 × 15 = 105°

4. Convert to degrees and minutes: (a) 0.0125 rad (b) 2.69 rad (c) 7.241 rad

(a) 0.0125 rad = 1800.0125π°

× = 0.716° or 0°43′

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(b) 2.69 rad = 1802.69π°

× = 154.126° or 154°8′

(c) 7.241 rad = 1807.241π°

× = 414.879° or 414°53′

5. A car engine speed is 1000 rev/min. Convert this speed into rad/s.

1000 rev/min = 1000 rev/min 2 rad/rev60s/min

π× = 104.7 rad/s

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EXERCISE 120 Page 280

1. Calculate the area of a circle of radius 6.0 cm, correct to the nearest square centimetre. Area of circle = ( )22 6.0rπ π= = 113.1 2cm 2. The diameter of a circle is 55.0 mm. Determine its area, correct to the nearest square millimetre.

Area of circle = 2 2

2 55.04 4

drπ π π = =

= 2376 2mm

3. The perimeter of a circle is 150 mm. Find its area, correct to the nearest square millimetre.

Perimeter = circumference = 150 = 2πr from which, radius, r = 150 752π π

=

Area of circle = 2

2 75rπ ππ

=

= 1790 2mm

4. Find the area of the sector, correct to the nearest square millimetre, of a circle having a radius of 35 mm, with angle subtended at centre of 75°.

Area of sector = ( ) ( )2 275 35360 360

rθ π π= × = 802 2mm

5. An annulus has an outside diameter of 49.0 mm and an inside diameter of 15.0 mm. Find its area correct to 4 significant figures.

Area of annulus = ( ) ( )2 21 22 2 2 2 2 21 2 1 2 49.0 15.0

4 4 4 4d dr r d dπ ππ π π π− = − = − = − = 1709 2mm

6. Find the area, correct to the nearest square metre, of a 2 m wide path surrounding a circular plot of land 200 m in diameter.

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Area of path = ( ) ( )2 2 2 21 2 204 2004 4

d dπ π− = − = 1269 2m

7. A rectangular park measures 50 m by 40 m. A 3 m flower bed is made round the two longer sides and one short side. A circular fish pond of diameter 8.0 m is constructed in the centre of the park. It is planned to grass the remaining area. Find, correct to the nearest square metre, the area of grass.

Area of grass = (50 × 40) – 2(50 × 3) – (34 × 3) – 28.0

4π = 2000 – 300 – 102 – 16π = 1548 2m

8. With reference to the diagram, determine (a) the perimeter and (b) the area.

(a) Perimeter = 17 + 28 + 17 + 12

(2π × 14) = 106.0 cm

(b) Area = (28 × 17) + ( )21 142π × = 476 + 98π = 783.9 2cm

9. Find the area of the shaded portion shown.

Shaded area = (10 × 10) – π(5) 2 = 100 – 25π = 21.46 2m

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10. Find the length of an arc of a circle of radius 8.32 cm when the angle subtended at the centre is 2.14 radians. Calculate also the area of the minor sector formed. Arc length, s = rθ = (8.32)(2.14) = 17.80 cm

Area of minor sector = ( ) ( )221 1 8.32 2.142 2

r θ = = 74.07 2m

11. If the angle subtended at the centre of a circle of diameter 82 mm is 1.46 rad, find the lengths of the (a) minor arc, and (b) major arc.

If diameter d = 82 mm, radius r = 822

= 41 mm

(a) Minor arc length, s = rθ = (41)(1.46) = 59.86 mm (b) Major arc length = circumference – minor arc = 2π(41) – 59.86 = 257.61 – 59.86 = 197.8 mm 12. A pendulum of length 1.5 m swings through an angle of 10° in a single swing. Find, in centimetres, the length of the arc traced by the pendulum bob.

Arc length of pendulum bob, s = rθ = (1.5) 10180π ×

= 0.262 m or 26.2 cm

13. Determine the shaded area of the section shown.

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Shaded area = (12 × 15) + 12

[π(8) 2 ] – π(5) 2 = 180 + 32π – 25π = 180 + 7π = 202 2mm

14. Determine the length of the radius and circumference of a circle if an arc length of 32.6 cm subtends an angle of 3.76 radians.

Arc length, s = rθ from which, radius, r = 32.63.76

sθ= = 8.67 cm

Circumference = 2πr = 2π(8.67) = 54.48 cm 15. Determine the angle of lap, in degrees and minutes, if 180 mm of a belt drive is in contact with a pulley of diameter 250 mm.

Arc length, s = 180 mm, radius, r = 2502

= 125 mm

Since s = rθ, the angle of lap, θ = 180125

sr= = 1.44 rad = 1801.44

π× = 82.5°

16. Determine the number of complete revolutions a motorcycle wheel will make in travelling 2 km, if the wheel’s diameter is 85.1 cm. If wheel diameter = 85.1 cm, then circumference, c = πd = π(85.1) cm = 267.35 cm = 2.6735 m

Hence, number of revolutions of wheel in travelling 2000 m = 20002.6735

= 748.08

Thus, number of complete revolutions = 748 17. The floodlights at a sports ground spread its illumination over an angle of 40° to a distance of 48 m. Determine (a) the angle in radians and (b) the maximum area that is floodlit.

(a) In radians, 40° = 40 rad180π

× = 0.69813 rad = 0.698 rad, correct to 3 decimal places

(b) Maximum area floodlit = area of sector = ( ) ( )221 1 48 0.698132 2

r θ = = 804.2 2m

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18. Find the area swept out in 50 minutes by the minute hand of a large floral clock, if the

hand is 2 m long.

Area swept out = ( )2250 50 260 60

rπ π× = × = 10.47 2m

19. Determine (a) the shaded area below and (b) the percentage of the whole sector that the area of the shaded area represents.

(a) Shaded area = 2 21 1(50) (0.75) (38) (0.75)2 2

− = [ ]2 21 (0.75) 50 382

− = 396 2mm

(b) Percentage of whole sector = 2

396 100%1 (50) (0.75)2

× = 42.24%

20. Determine the length of steel strip required to make the clip shown.

Angle of sector = 360° – 130° = 230° = 230180π

× rad = 4.01426 rad

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Thus, arc length, s = rθ = (125)(4.01426) = 501.783 mm Length of steel strip in clip = 100 + 501.783 + 100 = 701.8 mm 21. A 50° tapered hole is checked with a 40 mm diameter ball as shown below. Determine the length shown as x.

From the sketch below, tan 25° = 35AC

from which, AC = 35tan 25°

= 75.06 mm

and sin 25° = 20AB

from which, AB = 20sin 25°

= 47.32 mm

i.e. AC = 75.06 = x + 20 + AB = x + 20 + 47.32 and x = 75.06 – 20 – 47.32 i.e. x = 7.74 mm

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EXERCISE 121 Page 283

1. Determine (a) the radius, and (b) the coordinates of the centre of the circle given by the equation: x 2 + y 2 – 6x + 8y + 21 = 0 Method 1: The general equation of a circle 2 2 2( ) ( )x a y b r− + − = is 2 2 2 2 0x y ex fy c+ + + + =

where coordinate a = – 22e , coordinate b = – 2

2f and radius r = 2 2a b c+ −

Hence, if 2 2 6 8 21 0x y x y+ − + + = then a = – 2 62 2e −= − = 3, b = – 2 8

2 2f= − = –4

and radius, r = [ ]2 2(3) ( 4) (21) (9 16 21) 4+ − − = + − = = 2 i.e. the circle 2 2 6 8 21 0x y x y+ − + + = has (a) radius 2 and (b) centre at (3, –4), as shown below.

Method 2: 2 2 6 8 21 0x y x y+ − + + = may be rearranged as: 2 2( 3) ( 4) 4 0x y− + + − = i.e. 2 2 2( 3) ( 4) 2x y− + + = which has a radius of 2 and centre at (3, –4) 2. Sketch the circle given by the equation: x 2 + y 2 – 6x + 4y – 3 = 0 Method 1: The general equation of a circle 2 2 2( ) ( )x a y b r− + − = is 2 2 2 2 0x y ex fy c+ + + + =

where coordinate a = – 22e , coordinate b = – 2

2f and radius r = 2 2a b c+ −

Hence, if 2 2 6 4 3 0x y x y+ − + − = then a = – 2 62 2e −= − = 3, b = – 2 4

2 2f= − = –2

and radius, r = [ ]2 2(3) ( 2) ( 3) (9 4 3) 16+ − − − = + + = = 4

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i.e. the circle 2 2 6 4 3 0x y x y+ − + − = has centre at (3, –2) and radius 4, as shown below.

Method 2: 2 2 6 4 3 0x y x y+ − + − = may be rearranged as: 2 2( 3) ( 2) 16 0x y− + + − = i.e. 2 2 2( 3) ( 2) 4x y− + + = which has a radius of 4 and centre at (3, –2) 3. Sketch the curve: x 2 + (y – 1) 2 – 25 = 0 x 2 + (y – 1) 2 – 25 = 0 i.e. x 2 + (y – 1) 2 = 25 i.e. x 2 + (y – 1) 2 = 5 2 which represents a circle, centre (0, 1) and radius 5 as shown in the sketch below.

4. Sketch the curve: x = 62

16y −

If 2

6 16yx

= −

then 2

16 6x y = −

and

2 2

16 6x y = −

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i.e. 2 2

2 21

6 6x y

+ = and 2 2 26x y+ =

which is a circle of radius 6 and co-ordinates of centre at (0, 0), as shown below.

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EXERCISE 122 Page 284

1. A pulley driving a belt has a diameter of 300 mm and is turning at 2700/π revolutions per minute. Find the angular velocity of the pulley and the linear velocity of the belt, assuming that no slip occurs.

Angular velocity, ω = 2πn = 2π 270060π

= 90 rad/s

Linear velocity v = ωr = (90)3300 10

2

−×

= 13.5 m/s

2. A bicycle is travelling at 36 km/h and the diameter of the wheels of the bicycle is 500 mm, Determine the angular velocity of the wheels of the bicycle and the linear velocity of a point on the rim of one of the wheels.

Linear velocity, v = 36 km/h = 36 10003600× m/s = 10 m/s

(Note that changing from km/h to m/s involves dividing by 3.6)

Radius of wheel, r = 5002

= 250 mm = 0.25 m

Since, v = ωr, then angular velocity, ω = 100.25

vr= = 40 rad/s

3. A train is travelling at 108 km/h and has wheels of diameter 800 mm. (a) Determine the angular velocity of the wheels in both rad/s and rev/min. (b) If the speed remains constant for 2.70 km, determine the number of revolutions made by a wheel, assuming no slipping occurs.

(a) Linear velocity, v = 108 km/h = 1083.6

m/s = 30 m/s

Radius of wheel = 8002

= 400 mm = 0.4 m

Since, v = ωr, then angular velocity, ω = 300.4

vr= = 75 rad/s

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rev 60s75rad/s 75rad/s2 rad minπ

= × × = 716.2 rev/min

(b) Linear velocity, v = st

hence, time, t = 2700m30m/s

sv= = 90 s = 90

60 = 1.5 minutes

Since a wheel is rotating at 716.2 rev/min, then in 1.5 minutes it makes 716.2 rev/min × 1.5 min = 1074 rev/min

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EXERCISE 123 Page 286

1. Calculate the tension in a string when it is used to whirl a stone of mass 200 g round in a horizontal circle of radius 90 cm with a constant speed of 3 m/s.

Tension in string = centripetal force = 2 20.20kg (3m/s)

0.90mmv

r×

= = 2 N

2. Calculate the centripetal force acting on a vehicle of mass 1 tonne when travelling around a bend of radius 125 m at 40 km/h. If this force should not exceed 750 N, determine the reduction in speed of the vehicle to meet this requirement.

Centripetal force = 2mv

r where mass, m = 1 tonne = 1000 kg, radius, r = 125 m and

velocity, v = 40 km/h = 403.6

m/s

Hence, centrifugal force =

240(1000)3.6

125

= 988 N

If centrifugal force is limited to 750 N, then 2(1000)750

125v

=

from which, velocity, v = (750)(125)1000

= 9.6825 m/s

= 9.6825 × 3.6 = 34.89 km/h

Hence, reduction in speed is 40 – 34.89 = 5.14 km/h

3. A speed-boat negotiates an S-bend consisting of two circular arcs of radii 100 m and 150 m. If the speed of the boat is constant at 34 km/h, determine the change in acceleration when leaving one arc and entering the other.

Speed of the boat, v = 34 km/h = 343.6

m/s

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For the first bend of radius 100 m, acceleration =

2

22

1

343.6 0.892m/s100

vr

= =

For the second bend of radius 150 m, acceleration =

2

2

343.6 0.595m/s150

= − = − , the negative sign

indicating a change in direction Hence, change in acceleration is 0.892 – (–0.595) = 1.49 2m/s