1. Circles

2. Finding the radius of sectors Area of Segments Volumes of Cylinders Pi
Circle words
Rounding Refresher
Circumference
Area
Perimeter and Area of compound shapes
Perimeters of sectors
Area of Sectors
Finding the radius of sectors
Area of Segments
Volumes of Cylinders
Volume of Spheres and cones
Radius and Height of Cylinders
Equation of a circle 1
Equation of a circle 2
Simultaneous Equations
Circle theorems
Circle formulae

3. Match the words to the definitions
Sector
Segment
Chord
Radius
Arc
Tangent
Diameter
Circumference
The length around the outside of a circle
A line which just touches a circle at one point
A section of a circle which looks like a slice of pizza
A section circle formed with an arc and a chord
The distance from the centre of a circle to the edge
The distance from one side of a circle to the other (through the centre)
A section of the curved surface of a circle
A straight line connecting two points on the edge of a circle
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4. Think about circles Think about a line around the outside of a circle
Image that line straightened out- this is the circumference

5. You can use the π button on your calculatorPi
People noticed that if you divide the circumference of a circle by the diameter you ALWAYS get the same answer
They called the answer Pi (π) , which is:
You can use the π button on your calculator

6. How many digits can you memorise in 2 minutes?

7. How did you do? What do you think the world record is?
Write down pi!
How did you do? What do you think the world record is?

8. Pi website

9. Pi Story
One way to memorise Pi is to write a Pi-em (pi poem) where the number of letters in each word is the same as the number in pi. For example:
“Now I, even I, would celebrate in rhymes inept, the great immortal Syracusan rivall'd nevermore who in his wondrous lore passed on before left men his guidance how to circles mensurate.”
Can you write one of your own?
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10. Rounding to Decimal Places
10 multiple choice questions

11. Round to 1 dp
0.34
0.4
0.3 A) B)
0.35 3 C) D)

12. Round to 1 dp
0.48
0.49
0.4 A) B)
0.47
0.5 C) D)

13. Round to 1 dp
2.75
2.8
2.74 A) B)
2.7
3.0 C) D)

14. Round to 1 dp
13.374
13.0
14.0 A) B)
13.3
13.4 C) D)

15. Round to 1 dp
26.519
25.0
26.6 A) B)
26.0
26.5 C) D)

16. Round to 2 dp
23.20
23.18 A) B)
23.10
23.17 C) D)

17. Round to 2 dp
500.80
500.8 A) B)
500.83
500.84 C) D)

18. Round to 3 dp
0.004 A) B)
0.005 C) D)

19. Round to 2 dp
4.999
4.99
4.90 A) B)
5.00
4.98 C) D)

20. Round to 4 dp
0.7300
0.7390 A) B)
0.7210
0.7399 C) D)
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21. Finding the Circumference
You can find the circumference of a circle by using the formula-
Circumference = π x diameter
For Example-
Area= π x 10 =
= 31.4 cm (to 1 dp)
10cm

22. Circumference = π x diameter
You can find the circumference of a circle by using the formula-
Circumference = π x diameter
For Example-
Area= π x 10 =
= 31.4 cm (to 1 dp)
ANSWERS
10cm 1a
25.1cm b
12.6cm c
34.6cm d
66.0cm e
47.1cm
Find the Circumference of a circles with:
A diameter of :
8cm
4cm
11cm
21cm
15cm
A radius of :
6cm
32cm
18cm
24cm
50cm 2a
37.7cm b
201.1cm c
113.1cm d
150.8cm e
157.1cm
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23. Finding the Area Area= π x Radius2 For Example-
You can find the area of a circle by using the formula-
Area= π x Radius2
For Example-
Area= π x 72
= π x 49 =
= (to 1dp) cm2
7cm

24. Finding the Area Area= π x Radius2 2a 12.6 b 78.5 c 15.2 d 380.1 e
You can find the area of a circle by using the formula-
Area= π x Radius2
For Example-
Area= π x 72
= π x 49 =
= (to 1dp) cm2
7cm
ANSWERS 2a
12.6 b
78.5 c
15.2 d
380.1 e
314.2 f
153.9 g
100.5 h
28.3
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25. Finding the Area of a Sector
To find the area of a sector, you need to work out what fraction of a full circle you have, then work out the area of the full circle and find the fraction of that area.
For Example-
The sector here is ¾ of a full circle
Find the area of the full circle
Area= π x 72
= π x 49 =
= DON’T ROUND YET!
Then find ¾ of that area
¾ of = (divide by 4 and multiply by 3)
7cm

26. Finding the Area of a Sector
Sometimes it is not easy to see what fraction of a full circle you have.
You can work it out based on the size of the angle. If a full circle is 360°
, and this sector is 216°, the sector is 216/360, which can be simplified to 3/5.
For Example-
The sector here is 3/5 of a full circle
Find the area of the full circle
Area= π x 72
= π x 49 =
= DON’T ROUND YET!
Then find 3/5 of that area
3/5 of = (divide by 5 and multiply by 3)
= 92.4cm2
216°
7cm
Sometimes the fraction cannot be simplified and will stay over 360

27. Finding the Area of a Sector
The general formula for finding the area is:
Area of sector= Angle of Sector x πr2
360
Fraction of full circle that sector covers
“of”
Area of full circle

28. Find the area of these sectors, to 1 decimal place
Questions
Find the area of these sectors, to 1 decimal place 3
ANSWERS 1 2
10cm
11cm 1
226.9 2
200.6 3
315.4
12cm
260°
190°
251° 4
19.0 5
61.2 6
80.7 6 4 5
32°
5cm
6.5cm
17cm
87°
166°
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29. Finding the Perimeter of a Sector
To find the perimeter of a sector, you need to work out what fraction of a full circle you have, then work out the circumference of the full circle and find the fraction of that circumference.
You then need to add on the radius twice, as so far you have worked out the length of the curved edge
For Example-
The sector here is ¾ of a full circle
Find the area of the full circle
Area= π x 14 (the diameter is twice the radius)
= π x 49 =
= DON’T ROUND YET!
Then find ¾ of that circumference
¾ of = cm (2 dp)
Remember to add on 7 twice from the straight sides
7cm

30. Finding the Area of a Sector
Sometimes you will not be able to see easily what fraction of the full circle you have.
To find the fraction you put the angle of the sector over 360
This sector is 250/360 or two hundred and fifty, three hundred and sixty-ITHS of the full circle
Simplify if you can
250°
Sometimes the fraction cannot be simplified and will stay over 360

31. Finding the Perimeter of a Sector
The general formula for finding the area is:
Perimeter of sector= (Angle of Sector x πd) + r + r
360
Fraction of full circle that sector covers
“of”
Circumference of full circle
Don’t forget the straight sides
This is the same as d of 2r, but I like r +r as it helps me remember why we do it

32. Find the perimeter of these sectors, to 1 decimal place
Questions
Find the perimeter of these sectors, to 1 decimal place
ANSWERS 1
65.4 2
58.5 3
76.6 3 1 2
10cm
11cm
12cm
260°
190°
251° 4
17.6 5
31.8 6
43.5 6 4 5
32°
5cm
6.5cm
17cm
87°
166°
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33. Compound Area and Perimeter
Here we will look at shapes made up of triangles, rectangles, semi and quarter circles.
Find the area of the shape below:
10cm
Area of this semi circle = π r2 ÷ 2
= π x 52 ÷ 2
= π x 25 ÷ 2
=39.3 cm2 (1dp)
8cm
10cm
Area of this rectangle= 8 x10
=80cm2
Area of whole shape =
= cm2

34. Compound Area and Perimeter
Find the perimeter of the shape below:
10cm
Circumference of this semi circle = πd ÷ 2
= π x 10 ÷ 2
=15.7 cm (1dp)
8cm
10cm
Perimeter of this rectangle=
=26cm
(don’t include the red side)
Perimeter of whole shape =
= 31.7 cm

35. Compound Area and Perimeter
Find the areaof the shape below:
5cm
Area of this quarter circle = π r2 ÷ 4
= π x 52 ÷ 4
= π x 25 ÷ 4
=19.7 cm2 (1dp)
10cm
11cm
Area of whole shape =
= 129.7cm2
Area of this rectangle
10 x 11=110

36. Compound Area and Perimeter
Work out all missing sides first
Find the perimeter of the shape below: ?
5cm
Circumference of this quarter circle = πd ÷ 4
= π x 10 ÷ 4 (if radius is 5, diameter is 10)
=7.9 cm (1dp)
5cm
6cm
10cm
10cm
11cm
Area of whole shape =
= 49.9cm
Add all the straight sides=
= 42cm

37. Find the perimeter and area of these shapes, to 1 decimal place
Questions
ANSWERS
AREA
PERIMETER 1
38.1
23.4 2
135.0
61.3 3
181.1
60.8
Find the perimeter and area of these shapes, to 1 decimal place
2cm 3 1 2 4
27.3 5
129.3
47.7 6
128.5
12cm
10cm
20cm
6cm
11cm
4cm
17cm
4cm
6cm
Do not worry about perimeter here
Do not worry about perimeter here 6 4 5
10cm
10cm
5cm
5cm
5cm
12cm
20cm
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38. Volume of Cylinders Here we will find the volume of cylinders
Cylinders are prisms with a circular cross sections, there are two steps to find the volume
1) Find the area of the circle
1) Multiple the area of the circle by the height or length of the cylinder

39. Volume of Cylinders 2 EXAMPLE- find the volume of this cylinder 4cm
Find the area of the circle
π x r2
π x 42
π x 16 = 50.3 cm2 (1dp)
10cm
2) Multiple the area of the circle by the height or length of the cylinder
50.3 (use unrounded answer from calculator) x 10 = 503cm3

40. Find the volume of these cylinders, to 1 decimal place
Questions
Find the volume of these cylinders, to 1 decimal place 3
ANSWERS 1
4cm 2
3cm
5cm 1
603.2 2
282.7 3
1178.1
12cm
10cm
15cm 4
142.0 5
2155.1 6
508.9 6 4 5
3cm
2cm
7cm
18cm
11.3cm
14cm
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41. Volume of Cylinders 2 EXAMPLE- find the height of this cylinder 4cm
Find the area of the circle
π x r2
π x 42
π x 16 = 50.3 cm2 (1dp) h
2) Multiple the area of the circle by the height or length of the cylinder
50.3 x h = 140cm3
Rearrange this to give
h= 140 ÷ 50.3
h=2.8 cm
Volume= 140cm3

42. Volume of Cylinders EXAMPLE- find the radius of this cylinder r
Find the area of the circle
π x r2
2) Multiple the area of the circle by the height or length of the cylinder
π x r2 x 30 = 250cm3
x r2 = 250
Rearrange this to give
r2 = 250 ÷ 94.2
r2 =2.7 (1dp)
r= 1.6 (1dp) cm
30cm
Volume= 250cm3

43. Find the volume of these cylinders, to 1 decimal place
Questions
ANSWERS 1
6.4 2
4.2 3
1.3
Find the volume of these cylinders, to 1 decimal place 4
2.3 5
1.8 6
1.9 3 1
4cm 2
3cm
5cm h h h
volume= 320cm3
volume= 120cm3
volume= 100cm3 r 5 6 r 4 r
12cm
8cm
14cm
volume= 90cm3
volume= 200cm3
volume= 150cm3
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44. Volume of Spheres The formula for the volume of a sphere is e.g 10cm
A= 4/3 x π x 103
A= 4/3 x π x 1000
A= cm3 (1 dp)
10cm

45. Volume of Cones The formula for the volume of a cone is 10cm e.g
A= 1/3 x π x 42 x 10
A= 1/3 x π x 16 x 10
A=167.6 cm3 (1 dp)
4cm

46. Find the volume of these spheres, to 1 decimal place
Questions
ANSWERS 1
4188.8 2 3
523.6
Find the volume of these spheres, to 1 decimal place 4
201.1 5
122.5 6
1272.3 3 1 2
10cm
20cm
5cm 6
15cm
9cm 4
12cm
4cm 5
13cm
3cm
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47. Circles Theorems Angles connected by a chord Angle at the centre
Triangles made with a diameter or radii
Cyclic Quadrilaterals
Tangents

48. Double Angle x Angle x = 50 x 2 x=100°
The angle at the centre of a circle is twice the angle at the edge
50° x
Example
Angle x = 50 x 2
x=100°

49. 2 3 1
25°
60°
90°
Answers
1) 50
2)120
3)180 x x x
4)50
5)67.5
6)80 5 6 4 x x x
160°
100°
135°
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50. Triangles inside circles
A triangle containing a diameter, will be a right angled triangle
A triangle containing two radii, will be isosceles x
90°

51. 2 3 1 Answers 1) X=30 2)x=18 3)x=45 4)X=40 y=40 5)x=30 y= 120
72° x x
60° x
Answers
1) X=30
2)x=18
3)x=45 x x
4)X=40 y=40
5)x=30 y= 120
6)x=22 y=136 3 1 2 x y x
22°
100° y y
30° x
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52. Angles connected by a chord
Angles connected by a chord are equal x x y y

53. 2 3 1 Answers 1) x=25 y=15 2)x=125 y= 40 z=15 3)x=10 y=70 z=100
25° y
10° x x
40° z x
100°
125° y
15° z
15°
Answers
1) x=25 y=15
2)x=125 y= 40 z=15
3)x=10 y=70 z=100 y
4)X=105 y=40 z=35
5)x=53 y= 30 z=72
6)x=85 y=80 z=17 5 4 z y x
80°
17°
95° 6 z y y
40° x x z
35°
25°
53°
30°
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54. Tangents to a circle
A tangent will always meet a radius at 90°
90°

55. 1 x
35 ° 1
140° x 2
120° x 4
40° x y z 3 z y
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56. Cyclic Quadrilaterals
Opposite angles in a cyclic quadrilaterals add up to 180°
100 + y = 180
y=80°
60 + x = 180
x = 120 ° x y
100°
60°

57. 2 1 3 Answers 1) x=70 y=85 2)x=126 y=105 3)x=100 y=160 4 5
95°
110°
54°
20° x y x
75° y
80°
Answers
1) x=70 y=85
2)x=126 y=105
3)x=100 y=160 4 5
25° a
15°
70° w
4)w=15 x=70 y=65 z= 25
5)a=60 b=36 4b y z 2a x b
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58. Area of Segments Here we will look at finding the area of sectors
You will need to be able to do two things:
Find the area of a triangle using the formula-
Area= ½ absinC
Find the area of a sector using the formula-
Area of sector= Angle of Sector x πr2
360 a C b

59. Example- find the area of the blue segment
Step 1- find the area of the whole sector
Area= 100/360 x π x r2
= 100/360 x π x 102
=100/360 x π x 100
=87.3cm2
Step 2- find the area of the triangle
Area= ½ absinC
=1/2 x 10 x 10 x sin100
= 49.2cm2
10cm
10cm
100°
Step 3- take the area of the triangle from the area of the segment
87.3 – 49.3 = 38 cm2

60. Example- find the area of the blue segment
Step 1- find the area of the whole sector
Area= 120/360 x π x r2
= 120/360 x π x 122
=120/360 x π x 144
=150.8cm2
Step 2- find the area of the triangle
Area= ½ absinC
=1/2 x 12 x 12 x sin120
= 62.4cm2
12cm
12cm
120°
Step 3- take the area of the triangle from the area of the segment
150.8 – 62.4 = 88.4 cm2

61. Find the area of the blue segments, to 1 decimal place
Questions
ANSWERS 1
75.1 2
29.5 3
201.1
Find the area of the blue segments, to 1 decimal place 4
8.3 5
51.8 6
33.0 3 1 2
10cm
85°
11cm
170°
130°
12cm 6 4 5
6.5cm
95°
5cm
65°
17cm
160°
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62. Finding the Radius or angle of a Sector
10cm
100° x
Area=200cm2
Area=150
Area= 100 x π x r2
360
200= 100 x π x r2
200x360 = r2
100 x π
229.2=r2
15.1cm =r
Area= θ x π x r2
360
150= θ x π x 102
150x360 = θ
102 x π
117.9°= θ

63. Find the missing radii and angles of these sectors, to 1 decimal place
Questions
ANSWERS 1
7.6 2
8.3 3
5.4
Find the missing radii and angles of these sectors, to 1 decimal place 4
160.4 5
122.1 6
47.6 3 1 2 r r r
200°
175°
250°
Area=100cm2
Area=120cm2
Area=50cm2 6 4 5
5cm
6.5cm
17cm θ θ θ
Area=120cm2
Area=35cm2
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Area=45cm2

64. The Equation of a Circle
The general equation for a circle is (x-a)2 + (y-b)2=r2
This equation will give a circle whose centre is at (a,b) and has a radius of r
For example a circle has the equation (x-2)2 + (y-3)2=52
This equation will give a circle whose centre is at (2,3) and has a radius of 5

65. The Equation of a Circle
A circle has the equation (x-5)2 + (y-7)2=16
This equation will give a circle whose centre is at (5,7) and has a radius of 4 (square root of 16 is 4)
For example a circle has the equation (x+2)2 + (y-4)2=100
This equation will give a circle whose centre is at (-2,4) and has a radius of 10
You could think of this as (x - -2)2

66. The Equation of a Circle
A circle has the equation (x-5)2 + (y-7)2=16
What is y when x is 1?
(1-5)2 + (y-7)2=16
12+ (y-7)2=16
1+ (y-7)2=16
(y-7)2=15
y-7= ±3.9 (square root of 15 to 1 dp)
y= 7±3.9
y= 10.9 or 3.1
There are two coordinates on the circle with x=1, one is (1,10.9) and the other is (1,3.1)

67. The Equation of a Circle
1) Write down the coordinates of the centre point and radius of each of these circles:
(x-5)2 + (y-7)2=16
(x-3)2 + (y-8)2=36
(x+2)2 + (y-5)2=100
(x+2)2 + (y+5)2=49
(x-6)2 + (y+4)2=144
x2 + y2=4
x2 + (y+4)2=121
(x-1)2 + (y+14)2 -16=0
(x-5)2 + (y-9)2 -10=15
2) What is the diameter of a circle with the equation (x-1)2 + (y+3)2 =64
3) Calculate the area and circumference of the circle with the equation (x-5)2 + (y-7)2=16
4) Calculate the area and perimeter of the circle with the equation (x-3)2 + (y-5)2=16
5) Compare your answers to question 3 and 4, what do you notice, can you explain this?
6 ) A circle has the equation (x+2)2 + (y-4)2=100, find:
a) x when y=7
b) y when x=6
Answers
1a) r=4 centre (5,7)
b) r=6 centre (3,8)
c) r=4 centre (-2,5)
d) r=10 centre (-2,-5)
e) r=7 centre (6,-4)
f) r=12 centre (0,0)
g) r=411centre (0,-4)
h) r=4 centre (1,-14)
i) r=5 centre (5,9)
Answers
2) 16
3)Circumference = 25.1 Area=50.3
4)Circumference = 25.1 Area=50.3
5) Circles have the same radius but different centres, they are translations
6a) x= 11.5 or -7.5
b) y=11.3 or -3.3
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68. The Equation of a Circle 2
Here we will look at rearranging equations to find properties of the circle they represent
Remember- The general equation for a circle is (x-a)2 + (y-b)2=r2
The skill you will need is called completing the square, you may have used it to solve quadratic equations

69. The Equation of a Circle 2
Example
x2 + y2 -6x – 8y =0
Create two brackets and put x in one and y in the other
(x ) 2 + (y ) 2 = 0
Half the coefficients of x and y and put them into the brackets, and then subtract those numbers squared
(x -3) 2 + (y - 4) 2 – = 0
Tidy this up
(x -3) 2 + (y - 4) 2 – 25= 0
(x -3) 2 + (y - 4) 2 = 25
This circle has a radius of 5 and centre of (3,4)

70. The Equation of a Circle 2
Example
x2 + y2 -10x – 4y- 7 =0
Create two brackets and put x in one and y in the other
(x ) 2 + (y ) 2 = 0
Half the coefficients of x and y and put them into the brackets, and then subtract those numbers squared
(x -5) 2 + (y - 2) 2 – 52 – = 0
Tidy this up
(x -5) 2 + (y - 2) 2 – 36= 0
(x -5) 2 + (y - 2) 2 = 36
This circle has a radius of 6 and centre of (5,2)

71. The Equation of a Circle 2
You must always make sure the coefficient of x2 and y2 is 1
You may have to divide through 2x2 + 2y2 -20x – 8y- 14 =0
Divide by 2 to give x2 + y2 -10x – 4y- 7 =0
Then put into the form x2 + y2 -10x – 4y- 7 =0

72. Questions
Put this equations into the form (x-a)2 + (y-b)2=r2 then find the centre and radius of the circle
Answers
1) r=5 centre (4,2)
2) r=7 centre (6,3)
3) r=7 centre (2,5)
4) r=9 centre (5,7)
x2 + y2 -8x – 4y- 5 =0
x2 + y2 -12x – 6y- 4 =0
x2 + y2 -4x – 10y- 20 =0
x2 + y2 -10x – 14y- 7 =0
x2 + y2 -12x – 2y- 62 =0
2x2 +2y2 -20x – 20y- 28 =0
3x2 + 3y2 -42x – 24y- 36 =0
5x2 + 5y2 -100x – 30y- 60 =0
Answers
5) r=10 centre (6,1)
6) r=8 centre (5,5)
7) r=8 centre (6,4)
8) r=11 centre (10,3)
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73. Ways to solve quadratic equations-
Simultaneous Equations
A circle has the equation (x-5)2 + (y-7)2=16 and a line has an equation of y=2x+1, at what points does the line intercept the circle?
We need to substitute into the equation of the circle so that we only have x’s or y’s
Because y=2x +1 we can rewrite the equation of the circle but instead of putting “y” in we’ll write “2x+1”
So,
(x-5)2 + (2x-1-7)2=16
(x-5)2 + (2x-8)2=16 expand the brackets
x2-10x x2 – 32x +64 = simplify and make one side 0
5x2 -42x + 73=0 solve this quadratic equation to find x,
Put the value / values of x into y=2x+1 to find the coordinates of the intercept / intercepts to answer the question
Ways to solve quadratic equations-
Completing the square
Factorising
The Quadratic formula

74. Simultaneous Equations
A quadratic equation can give 1,2 or no solutions, a line can cross a circle at 1,2 or no points
1 solution to the quadratic-
The line is a tangent
2 solutions to the quadratic
0 solutions to the quadratic the circle and the line never meet

75. Intercepts between lines and circles
1) Find out whether these circles and lines intercept, if they do find the coordinates of the interceptions
(x-5)2 + (y-7)2=16 and y=3x-1
(x-3)2 + (y-8)2=36 and y=2x-2
(x+2)2 + (y-5)2=100 and y=3x + 3
(x+2)2 + (y+5)2=49 and 2y+4=x
(x-6)2 + (y+4)2=144 and y -3x =5
ANSWERS (all have been rounded)
(3.6,9.8)
and
(2.2,5.6)
(7.2,12.3)
(2,2)
(3.5,13.4)
(-2.7,-5)
(-0.4,3.3)
(-6.4,-8.9)
(-4.6,6.9)
(-4.6,-8.9)
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76. Circle Formulae
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