3. Naming the parts of a circle
A circle is a set of points equidistant from its centre.
circumference: distance around the outside of a circle
tangent
radius: distance from the centre of the circle to the circumference
diameter
radius
diameter: distance across the width of the circle through the centre
centre
chord: line segment with two endpoints on the circle
chord
tangent: straight line that touches the circle at a single point
circumference

4. Arcs and sectors An arc is a part of the circumference. arc
When an arc is bounded by two radii a sector is formed.
sector
A minor arc is shorter than a semicircle and a major arc is longer than a semicircle.
A segment is a region bounded by a chord and an arc lying between the chord's endpoints.
segment

5. Arcs and sectors

6. Tangents

7. Tangents

8. Tangents

9. Chords

10. The alternate segment theorem

12. Right angles in a semicircle
Teacher notes
Drag the points around the circumference of the circle to demonstrate that angle ABC is always a right angle.

13. Right angles in a semicircle
Teacher notes
Stress the difference between a mathematical proof and a demonstration.
Explain that each vertex and the centre of the circle should be labelled. This is so that angles and line segments can be referred to using these letters.
The centre of a circle is usually called O.
State the importance of giving a reason for each statement. These reasons are shown in brackets. Explaining a proof in this way displays excellent quality of written communication.

14. Calculating the size of unknown angles
Teacher notes
Stress that each time the size of an angle is given a reason must be included. Explain that in an examination, marks can be awarded for correct reasons even if the angle given is incorrect.
a = 37° (angles at the base of an isosceles triangle)
b = 90° – 37°
= 53° (angle in a semi-circle)
c = 53° (angles at the base of an isosceles triangle)
d = 180° – 2 × 53°
= 74° (angles in a triangle)
e = 180° – 74°
= 106° (angles on a line)

16. The angle at the centre Teacher notes
Drag the points around the circumference of the circle to demonstrate that the angle at the centre of the circle is always double the angle at the circumference.
Hide one of the angles, modify the diagram and ask pupils to calculate the size of the missing angle.
Be aware that the angles have been given to the nearest degree.

17. The angle at the centre
The angle at the centre of a circle is twice the angle at the circumference.
This result can be proven as follows.
Draw a line from B, through the centre O, and to the other side D. B
In triangle AOB, x y
OA = OB
(both radii) O y
So, angle OAB = angle OBA x C
(angles at base of isosceles triangle) A
These angles can be called x. D
Similarly, angle OBC = angle OCB

18. The angle at the centre
The angle at the centre of a circle is twice the angle at the circumference.
The exterior angle in a triangle is equal to the sum of the opposite interior angles. B x y
angle AOD = 2x
angle COD = 2y O y
angle AOC = 2x + 2y 2x 2y x C
= 2(x + y) A
angle ABC = x + y D
 angle AOC = 2 × angle ABC

19. Calculating the size of unknown angles
Teacher notes
Stress that each time the size of an angle is given a reason must be included.
Explain that in an examination marks can be awarded for correct reasons even if the angle given is incorrect.
a = 29° (angles at the base of an isosceles triangle)
b = 180° – 2 × 29°
= 122° (angles in a triangle)
c = 122° ÷ 2
= 61° (angle at the centre is twice angle on the circumference)
d = 180° – (29° + 29° + 41° + 61°)
= 20° (angles in a triangle)

21. Angles in the same segment
Teacher notes
Drag the points around the circumference of the circle to demonstrate this theorem.
Hide one of the angles, modify the diagram and ask pupils to calculate the size of the missing angle.

22. Angles in the same segment
The angles in the same segment are equal.
This can be proven as follows.
Mark the centre of the circle O and show angle AOB. C D
angle ADB = of angle AOB 1 2
angle ACB = of angle AOB 1 2 O
Teacher notes
Discuss the fact that we can prove this circle theorem using the previous circle theorem.
(the angle at the centre of a circle is twice the angle at the circumference) B A
 angle ADB = angle ACB

23. Calculating the size of unknown angles
Teacher notes
Stress that each time the size of an angle is given, a reason must be included. Stress that in an examination marks can be awarded for correct reasons even if the angle given is incorrect.
a = 90° – 51°
= 39° (angle in a semi-circle)
b = 180° – (90° + 44°)
= 46° (angles in a triangle)
c = b = 46° (angles in the same segment)
d = 51° (angles in the same segment)

25. Angles in a cyclic quadrilateral
Teacher notes
Drag the points around the circumference of the circle to demonstrate that opposite angles in a cyclic quadrilateral add up to 180°.
Hide some of the angles, modify the diagram and ask pupils to calculate the sizes of the missing angles.

26. Angles in a cyclic quadrilateral
The opposite angles in a cyclic quadrilateral add up to 180°.
The result can be proven as follows.
Mark the centre of the circle O and label angles ABC and ADC x and y. B x
The angles at the centre are 2x and 2y. 2y
(the angle at the centre of a circle is twice the angle at the circumference) O 2x C A
2x + 2y = 360° y
2(x + y) = 360° D
x + y = 180°

27. Angles in a cyclic quadrilateral
If the opposite angles of a quadrilateral add up to 180°, a circle can be drawn through each of its vertices.
This is a cyclic quadrilateral since the opposite angles in this quadrilateral add up to 180°.
67°
109°
Remember that when two angles add up to 180° they are often called supplementary angles.
113°
71°

28. Calculating the size of unknown angles
Teacher notes
Stress that each time the size of an angle is given, a reason must be included. Stress that in an examination, marks can be awarded for correct reasons even if the angle given is incorrect.
a = 64° (angle at the centre)
b = c = (180° – 128°) ÷ 2
= 26° (angles at the base of an isosceles triangle)
d = 33° (angles at the base of an isosceles triangle)
e = 180° – 2 × 33°
= 114° (angles in a triangle)
f = 180° – (e + c)
= 180° – 140°
= 40° (opposite angles in a
cyclic quadrilateral)

30. The skaters’ dance
Two skaters practice their routine on a circular rink. They each start at one of the 2 points and skate across to another point on the circumference, then skate across to their partner’s starting position.
What can you say about the angles between their outward and return route?
Teacher notes
The angles are equal.
Where must they start skating so that the angle is 90°?
They should be at either end of the diameter. 30

31. The skaters’ dance
They now skate to the centre of the rink and return to their partner’s position. Then they skate across to a point on the circumference and return to their original position.
What is special about the angles between their outward and return routes?
Teacher notes
The angle at the centre of a circle is twice the angle at the circumference.
Where must they start skating so that the angle is 45° at the circumference?
This is one possible solution. The skaters must be a quarter of the circumference apart. 31

32. The skaters’ dance
This time the skaters can only use half of the rink. They skate to the circumference and return to their partner’s place.
What can you say about the angles
between their outward and return routes?
If there are 2 more skaters on the other half of the rink, where should they begin so that the angle is the same as the ones on the other side?
Teacher notes
Pupils might like to draw this problem to test their theories.
The angles in the same segment are equal.
The skaters must be the same distance apart and skate to a point in the major segment. 32