Circles Pi Circle words Rounding Refresher Area Perimeter

Circles

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

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 HOME

Think about circles Image about that line out- the this isoutside the circumference Think astraightened line around of a circle

Pi 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: 3. 1415926535897932384626433832795028841971693993751 058209749445923078164062862089986280348253421170679 821480865132823066470938446095505822317253594081284 8111745028410270193852110555964462294895493038196 You can use the π button 442881097566593344612847564823378678316527120190914 on your calculator 5648566923460348610454326648213393607260249141273 724587006606315588174881520920962829254091715364367 892590360011330530548820466521384146951941511609. . .

How many digits can you memorise in 2 minutes? 3. 141592653589793238462643383279502884197169399375 10582097494459230781640628620899862803482534211 70679821480865132823066470938446095505822317253 59408128481117450284102701938521105559644622948 95493038196442881097566593344612847564823378678 31652712019091456485669234603486104543266482133 93607260249141273724587006606315588174881520920 96282925409171536436789259036001133053054882046 6521384146951941511609. . .

Write down pi! 3. 141592653589793238462643383279502884197169399375 10582097494459230781640628620899862803482534211 70679821480865132823066470938446095505822317253 59408128481117450284102701938521105559644622948 95493038196442881097566593344612847564823378678 31652712019091456485669234603486104543266482133 93607260249141273724587006606315588174881520920 96282925409171536436789259036001133053054882046 6521384146951941511609. . . How did you do? What do you think the world record is?

Pi website

Finding the Circumference You can find the circumference of a circle by using the formula- Circumference = π x diameter For Example 10 cm Area= π x 10 = 31. 41592654. . = 31. 4 cm (to 1 dp)

You can find the circumference of a circle by using the formula- Circumference = π x diameter For Example- 10 cm Area= π x 10 = 31. 41592654. . = 31. 4 cm (to 1 dp) Find the Circumference of a circles with: 1. A diameter of : a) 8 cm b) 4 cm c) 11 cm d) 21 cm e) 15 cm 2. A radius of : a) 6 cm b) 32 cm c) 18 cm d) 24 cm e) 50 cm ANSWERS 1 a 25. 1 cm b 12. 6 cm c 34. 6 cm d 66. 0 cm e 47. 1 cm 2 a b c d e 37. 7 cm 201. 1 cm 113. 1 cm 150. 8 cm 157. 1 cm HOME

Finding the Area You can find the area of a circle by using the formula- Area= π x Radius 2 For Example 7 cm Area= π x 72 = π x 49 = 153. 93804 = 153. 9 (to 1 dp) cm 2

Finding the Area You can find the area of a circle by using the formula- Area= π x Radius 2 7 cm For Example. Area= π x 72 = π x 49 = 153. 93804 = 153. 9 (to 1 dp) cm 2 ANSWERS 2 a b c d e f g h HOME 12. 6 78. 5 15. 2 380. 1 314. 2 153. 9 100. 5 28. 3

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 7 cm Area= π x 72 = π x 49 = 153. 93804 = DON’T ROUND YET! Then find ¾ of that area ¾ of 153. 93804 = 115. 45353 (divide by 4 and multiply by 3)

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 216° 7 cm Area= π x 72 = π x 49 = 153. 93804 = DON’T ROUND YET! Then find 3/5 of that area 3/5 of 153. 93804 = 92. 362824 (divide by 5 and multiply by 3) = 92. 4 cm 2 Sometimes the fraction cannot be simplified and will stay over 360

Finding the Area of a Sector The general formula for finding the area is: Area of sector= Angle of Sector x πr 2 360 Fraction of full circle that sector covers “of” Area of full circle

Questions Find the area of these sectors, to 1 decimal place 1 3 2 10 cm 11 cm 260° 190° 4 12 cm 251° 6 5 87° ANSWERS 5 cm 6. 5 cm 32° 1 2 3 4 5 6 226. 9 200. 6 315. 4 19. 0 61. 2 80. 7 17 cm 166° HOME

Finding the Perimeter of a Sector (Arc Length) 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 7 cm Area= π x 14 (the diameter is twice the radius) = π x 49 = 43. 982297. . . = DON’T ROUND YET! Then find ¾ of that circumference ¾ of 43. 982297. . . = 32. 99 cm (2 dp) Remember to add on 7 twice from the straight sides

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 360 250° This sector is 250/360 or two hundred and fifty, three hundred and sixty-ITHS of the full circle Simplify if you can Sometimes the fraction cannot be simplified and will stay over 360

Finding the Perimeter of a Sector (Arc Length) The general formula for finding the area is: Perimeter of sector= (Angle of Sector x πd) + 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 2 r, but I like r +r as it helps me remember why we do it

Questions Find the perimeter of these sectors, to 1 decimal place 1 3 2 10 cm 11 cm 260° 190° 4 251° 6 5 87° 12 cm 5 cm 6. 5 cm 32° ANSWERS 1 65. 4 2 58. 5 3 76. 6 4 17. 6 5 31. 8 6 43. 5 17 cm 166° HOME

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: 10 cm 8 cm 10 cm Area of this rectangle= 8 x 10 =80 cm 2 Area of this semi circle = π r 2 ÷ 2 = π x 52 ÷ 2 = π x 25 ÷ 2 =39. 3 cm 2 (1 dp) Area of whole shape = 80 + 39. 3 = 119. 3 cm 2

Compound Area and Perimeter Find the perimeter of the shape below: 10 cm 8 cm 10 cm Perimeter of this rectangle= 8 + 10 =26 cm (don’t include the red side) Circumference of this semi circle = πd ÷ 2 = π x 10 ÷ 2 =15. 7 cm (1 dp) Perimeter of whole shape = 26 + 15. 7 = 31. 7 cm

Compound Area and Perimeter Find the areaof the shape below: Area of this quarter circle = π r 2 ÷ 4 = π x 52 ÷ 4 = π x 25 ÷ 4 =19. 7 cm 2 (1 dp) 5 cm 10 cm 11 cm Area of this rectangle 10 x 11=110 Area of whole shape = 110+ 19. 7 = 129. 7 cm 2

Compound Area and Perimeter Work out all missing Find the perimeter of the shape below: sides first ? Circumference of this quarter circle = πd ÷ 4 = π x 10 ÷ 4 (if radius is 5, diameter is 10) =7. 9 cm (1 dp) 5 cm 6 cm 10 cm 11 cm Add all the straight sides= 10+10 + 11+ 5 + 6= 42 cm Area of whole shape = 42+ 7. 9 = 49. 9 cm

Questions ANSWERS AREA PERIMETER 1 38. 1 23. 4 135. 0 61. 3 Find the perimeter and area of these shapes, to 2 1 decimal place 3 181. 1 60. 8 2 cm 3 4 27. 3 1 2 12 cm 10 cm 5 129. 3 47. 7 20 cm 6 128. 5 6 cm 11 cm 4 cm 17 cm 4 Do not worry about perimeter here 5 cm 12 cm 6 cm 5 6 Do not worry about perimeter here 10 cm 5 cm 10 cm 20 cm HOME

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

Volume of Cylinders 2 EXAMPLE- find the volume of this cylinder 1) Find the area of the circle π x r 2 π x 42 π x 16 = 50. 3 cm 2 (1 dp) 2) Multiple the area of the circle by the height or length of the cylinder 50. 3 (use unrounded answer from calculator) x 10 = 503 cm 3 4 cm 10 cm

Questions Find the volume of these cylinders, to 1 decimal place 1 2 4 cm 3 cm 12 cm 3 ANSWERS 5 cm 10 cm 15 cm 1 603. 2 2 282. 7 3 1178. 1 4 142. 0 5 2155. 1 4 2 cm 5 11. 3 cm 6 7 cm 14 cm 6 508. 9 3 cm 18 cm HOME

Volume of Cylinders 2 EXAMPLE- find the height of this cylinder 4 cm 1) Find the area of the circle π x r 2 π x 42 π x 16 = 50. 3 cm 2 (1 dp) h 2) Multiple the area of the circle by the height or length of the cylinder 50. 3 x h = 140 cm 3 Rearrange this to give h= 140 ÷ 50. 3 h=2. 8 cm Volume= 140 cm 3

Volume of Cylinders EXAMPLE- find the radius of this cylinder r 1) Find the area of the circle π x r 2 2) Multiple the area of the circle by the height or length of the cylinder π x r 2 x 30 = 250 cm 3 94. 2. . . x r 2 = 250 Rearrange this to give r 2 = 250 ÷ 94. 2 r 2 =2. 7 (1 dp) r= 1. 6 (1 dp) cm 30 cm Volume= 250 cm 3

Questions 1 ANSWERS 1 6. 4 2 4. 2 Find the volume of these cylinders, to 1 decimal place 3 1. 3 4 2. 3 3 4 cm 2 5 1. 8 3 cm 5 cm 6 1. 9 h h volume= 320 cm 3 4 volume= 120 cm 3 5 r volume= 100 cm 3 6 r 12 cm volume= 200 cm 3 h r 8 cm 14 cm volume= 150 cm 3 volume= 90 cm 3 HOME

Volume of Spheres The formula for the volume of a sphere is e. g A= 4/3 x π x 103 A= 4/3 x π x 1000 A=4188. 8 cm 3 (1 dp) 10 cm

Volume of Cones The formula for the volume of a cone is e. g A= 1/3 x π x 42 x 10 A= 1/3 x π x 16 x 10 A=167. 6 cm 3 (1 dp) 10 cm 4 cm

Questions 1 ANSWERS 1 4188. 8 2 33510. 3 3 523. 6 Find the volume of these spheres, to 1 decimal place 4 201. 1 3 5 122. 5 2 6 1272. 3 10 cm 4 20 cm 5 cm 6 5 12 cm 13 cm 4 cm 3 cm 15 cm 9 cm HOME

Circles Theorems Angle at the centre Angles connected by a chord Triangles made with a diameter or radii Cyclic Quadrilaterals Tangents

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

1 Answers 1) 50 2)120 3)180 4)50 5)67. 5 6)80 4 2 3 25° 60° 90° x x 100° 5 x 135° 6 x 160° HOME

Triangles inside circles A triangle containing a diameter, will be a right angled triangle A triangle containing two radii, will be isosceles 90° x x

2 1 3 72° x x 60° Answers x 1) X=30 2)x=18 3)x=45 4)X=40 y=40 5)x=30 y= 120 6)x=22 y=136 1 x x x y 3 2 x 22° 100° y 30° y x HOME

Angles connected by a chord are equal x x y y

2 1 25° x 15° Answers 1) x=25 y=15 2)x=125 y= 40 z=15 3)x=10 y=70 z=100 4)X=105 y=40 z=35 5)x=534 y= 30 z=72 6)x=85 y=80 z=17 y x 10° x 40° x y z 3 y z 100° 125° z 15° y 5 6 z y 40° 35° x 25° 30° 53° y x 17° 80° 95° z HOME

Tangents to a circle A tangent will always meet a radius at 90°

1 2 x 35 ° x 140° 4 z 3 x 40° z 120° x y y HOME

Cyclic Quadrilaterals Opposite angles in a cyclic quadrilaterals add up to 180° 100° 60° x y 100 + y = 180 y=80° 60 + x = 180 x = 120 °

2 1 95° 3 54° 110° 20° x y x Answers 1) x=70 y=85 2)x=126 y=105 3)x=100 y=160 4)w=15 x=70 y=65 z= 25 5)a=60 b=36 x 75° y 4 80° y 5 15° 25° 70° a w 4 b y z x b 2 a HOME

Area of Segments Here we will look at finding the area of sectors You will need to be able to do two things: 1) Find the area of a sector using the formula- 2) Find the area of a triangle using the formula. Area= ½ absin. C Area of sector= Angle of Sector x πr 2 360 a C b

Examplefind the area of the blue segment Step 1 - find the area of the whole sector Area= 100/360 x π x r 2 = 100/360 x π x 102 =100/360 x π x 100 =87. 3 cm 2 10 cm 100° 10 cm Step 2 - find the area of the triangle Area= ½ absin. C =1/2 x 10 x sin 100 = 49. 2 cm 2 Step 3 - take the area of the triangle from the area of the segment 87. 3 – 49. 3 = 38 cm 2

Examplefind the area of the blue segment Step 1 - find the area of the whole sector Area= 120/360 x π x r 2 = 120/360 x π x 122 =120/360 x π x 144 =150. 8 cm 2 12 cm 120° 12 cm Step 2 - find the area of the triangle Area= ½ absin. C =1/2 x 12 x sin 120 = 62. 4 cm 2 Step 3 - take the area of the triangle from the area of the segment 150. 8 – 62. 4 = 88. 4 cm 2

Questions 1 ANSWERS 1 75. 1 2 29. 5 3 201. 1 Find the area of the blue segments, to 1 decimal place 4 8. 3 3 2 5 51. 8 10 cm 6 33. 0 85° 11 cm 170° 12 cm 130° 4 6 5 95° 6. 5 cm 160° 65° 17 cm HOME

Finding the Radius or angle of a Sector r 100° Area=200 cm 2 Area= 100 x π x r 2 360 200 x 360 = r 2 100 x π 229. 2=r 2 15. 1 cm =r 10 cm x Area=150 Area= θ x π x r 2 360 150= θ x π x 102 360 150 x 360 = θ 102 x π 117. 9°= θ

Questions ANSWERS 1 7. 6 2 8. 3 Find the missing radii and angles of these sectors, to 1 decimal 3 place 5. 4 4 160. 4 3 1 2 5 122. 1 r r r 6 47. 6 200° 175° Area=100 cm 2 4 250° Area=120 cm 2 6 5 θ 5 cm Area=35 cm 2 6. 5 cm θ Area=45 cm 2 Area=50 cm 2 17 cm θ Area=120 cm 2 HOME

The Equation of a Circle The general equation for a circle is (x-a)2 + (y-b)2=r 2 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

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

The Equation of a Circle A circle has the equation (x-5)2 + (y-7)2=16 What is y when x is 10? (10 -5)2 + (y-7)2=16 5+ (y-7)2=16 (y-7)2=11 y-7= ± 3. 3 (square root of 11 to 1 dp) y= 7± 3. 3 y= 10. 3 or 3. 7 There are two coordinates on the circle with x=10, one is (10, 10. 3) and the other is (10, 3. 7)

The Equation of a Circle 1) Write down the coordinates of the centre point and radius of each of these circles: Answers a) (x-5)2 + (y-7)2=16 b) (x-3)2 + (y-8)2=36 c) (x+2)2 + (y-5)2=100 d) (x+2)2 + (y+5)2=49 e) (x-6)2 + (y+4)2=144 f) x 2 + y 2=4 g) x 2 + (y+4)2=121 h) (x-1)2 + (y+14)2 -16=0 i) (x-5)2 + (y-9)2 -10=15 1 a) 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=411 centre (0, -4) h) r=4 centre (1, -14) i) r=5 centre (5, 9) 2) What is the diameter of a circle with the equation (x-1)2 + (y+3)2 =64 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 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 6 a) x= 11. 5 or -7. 5 b) y=11. 3 or -3. 3 HOME

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=r 2 The skill you will need is called completing the square, you may have used it to solve quadratic equations

The Equation of a Circle 2 Example x 2 + y 2 -6 x – 8 y =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 – 32 - 42= 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)

The Equation of a Circle 2 Example x 2 + y 2 -10 x – 4 y- 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 – 22 - 7= 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)

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

Questions Put this equations into the form (x-a)2 + (y-b)2=r 2 then find the centre and radius of the circle 1. 2. 3. 4. 5. 6. 7. 8. x 2 + y 2 -8 x – 4 y- 5 =0 x 2 + y 2 -12 x – 6 y- 4 =0 x 2 + y 2 -4 x – 10 y- 20 =0 x 2 + y 2 -10 x – 14 y- 7 =0 x 2 + y 2 -12 x – 2 y- 62 =0 2 x 2 +2 y 2 -20 x – 20 y- 28 =0 3 x 2 + 3 y 2 -42 x – 24 y- 36 =0 5 x 2 + 5 y 2 -100 x – 30 y- 60 =0 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) 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) HOME

Simultaneous Equations A circle has the equation (x-5)2 + (y-7)2=16 and a line has an equation of y=2 x+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=2 x +1 we can rewrite the equation of the circle but instead of putting “y” in we’ll write “ 2 x+1” Ways to solve quadratic equations. Completing the square Factorising The Quadratic formula So, (x-5)2 + (2 x-1 -7)2=16 (x-5)2 + (2 x-8)2=16 expand the brackets x 2 -10 x + 25 + 4 x 2 – 32 x +64 = 16 simplify and make one side 0 5 x 2 -42 x + 73=0 solve this quadratic equation to find x, Put the value / values of x into y=2 x+1 to find the coordinates of the intercept / intercepts to answer the question

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

Intercepts between lines and circles 1) Find out whether these circles and lines intercept, if they do find the coordinates of the interceptions a) (x-5)2 + (y-7)2=16 b) (x-3)2 + (y-8)2=36 c) (x+2)2 + (y-5)2=100 and d) (x+2)2 + (y+5)2=49 e) (x-6)2 + (y+4)2=144 ANSWERS (all have been rounded) and y=2 x-2 (3. 6, 9. 8) and (2. 2, 5. 6) y=3 x + 3 (7. 2, 12. 3) and (2, 2) and 2 y+4=x (3. 5, 13. 4) and (-2. 7, -5) and y -3 x =5 (-0. 4, 3. 3) and (-6. 4, -8. 9) (-4. 6, 6. 9) and (-4. 6, -8. 9) and y=3 x-1 HOME

Circle Formulae HOME