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In order to access this I need to be confident with:

Parts of a circle Angles in polygons Angles on a straight line Angles around a point Angles in parallel linesProperties of triangles

FactorisingThis topic is relevant for:

Here we will learn about the circle theorem: the **angle at the centre,** including its application, proof, and using it to solve more difficult problems.

There are also circle theorem worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

The angle at the centre is twice the angle at the circumference of a circle.

In the diagram above, the two points B and D on the circumference are joined to the centre, C , and to another point on the circumference, A . The angle BCD is twice the size of the angle BAD . Therefore, **the angle at the centre is twice the angle at the circumference**.

There are multiple ways of viewing this theorem as it depends on the location of the point A on the circumference.

Below, we have moved point A on the circumference to show you how this theorem can appear in different circles.

- The
**centre**of the circle is a point that locates the middle of a circle. - The
**circumference**of the circle is the distance around the edge of the circle. - A
**chord**is a straight line that meets the circumference in two places. The longest chord in a circle is the**diameter**.

To be able to prove this theorem, you do not need to know any other circle theorem. You just need to be confident with **angles in a triangle** and **angles around a point**.

Step

1

2

3

4

5

6

7

8

9

10

11

12

Diagram

Description

A, B, and D are points on the

circumference and C is the centre

of the circle.

BC and CD are radii.

AB and AD are chords.

Connect the points AC so that you have

two triangles (ACD and ABC).

Let’s inspect triangle ACD.

We know that AC and CD are radii and

so this is an isosceles triangle. We can

therefore state the angles at A and D

are equal, so we have labelled them x.

As angles in a triangle add up to 180^o ,

the angle at C will be 180 degrees

minus the sum of the other two angles.

This is expressed as 180-2x.

Now let us inspect triangle ABC.

BC and AC are radii and so we have

another isosceles triangle. As we

cannot say that the angles at A and B

are the same as angle x in triangle

ACD, we say that both the angles are

equal to another number, y.

As angles in a triangle total 180^o , we

can state that the angle at C is

180-2y for the same reason as in

step 5.

Now we have the diagram with all the

angles filled in. We still need to prove

that the external angle to BCD is equal

to twice the angle at A but for now, let

us simplify each angle within ABCD.

The angle at A is now x+y and the

angle at C is the sum of 180-2x and

180-2y (or 180-2x+180-2y ).

As we need to know the angle on the

other side of BCD, we can use the fact

that angles around a point total 360^o.

\begin{aligned}
&360-(360-2x-2y) \\
&=360-360+2x+2y \\
&=2x+2y \\
\end{aligned}

We now have the external angle BCD

equalling 2x+2y.

If we factorise this expression, we get

our angle BCD as 2(x+y)

So as angle BAD =x+y and angle

BCD =2(x+y) , the angle at the centre

is twice the angle at the circumference.

In order to use the fact that the angle at the centre is twice the angle at the circumference

- Locate the key parts of the circle for the theorem
- Use other angle facts to determine the angle at the centre or the angle at the circumference
- Use the angle at the centre theorem to state the other missing angle

Get your free angle at the centre worksheet of 20+ questions and answers. Includes reasoning and applied questions.

COMING SOONGet your free angle at the centre worksheet of 20+ questions and answers. Includes reasoning and applied questions.

COMING SOONABCD is an arrowhead where C is the centre of the circle and A, B, and D lie on the circumference. Calculate the size of angle BAD .

**Locate the key parts of the circle for the theorem**.

Here we have:

- The angle BCD = 150^o
- BC = CD = Radii
- AB and AD are chords
- The angle BAD = \theta

2**Use other angle facts to determine the angle at the centre or the angle at the circumference**.

We already know that BCD = 150^o so we do not need to use any other angle fact to determine this angle for this example.

3** Use the angle at the centre theorem to state the other missing angle**.

The **angle at the centre is twice the angle at the circumference** and so as we know the angle at the centre, we need to divide this number by 2 to get the angle BAD :

BAD = 150 ÷ 2

BAD = 75°

Below is a circle with centre C . Points A, B, and D are on the circumference of the circle. Calculate the size of angle \theta.

**Locate the key parts of the circle for the theorem**.

Here we have:

- AC = BC = CD = Radii
- AB is a chord
- AD is the diameter
- The angle ABC = 46^o
- The angle BCD = \theta

**Use other angle facts to determine the angle at the centre or the angle at the circumference**.

As AC = BC , we can say that ABC is an isosceles triangle. This means that the angle BAC = angle ABC = 46^o .

** Use the angle at the centre theorem to state the other missing angle**.

The **angle at the centre is twice the angle at the circumference** and so as we know the angle at the circumference BAD , we need to multiply this number by 2 to get the angle BCD at the centre:

BCD = 46 × 2

BCD = 92°

ABCD is a cyclic quadrilateral around the centre E . Calculate the size of the reflex angle AEC , labelled \theta.

** Locate the key parts of the circle for the theorem**.

Here we have:

- AC = BC = CD = Radii
- AD and CD are chords
- The angle ADC = 72^o
- The reflex angle AEC = \theta

** Use other angle facts to determine the angle at the centre or the angle at the circumference**.

As ABCD is a cyclic quadrilateral, **opposite angles in a cyclic quadrilateral total ** 180^o . This means that angle ABC = 180 - 72 = 108^o . We therefore have the updated diagram:

**Use the angle at the centre theorem to state the other missing angle**.

The **angle at the centre is twice the angle at the circumference** and so as we know the angle at the circumference ABC , we need to multiply this number by 2 to get the angle AEC at the centre:

AEC = 108 × 2

AEC = 216°

A, B, and D are points on the circumference of the circle with centre C . Calculate the value of \theta.

**Locate the key parts of the circle for the theorem**.

Here we have:

- BC = CD = Radii
- AB, AD and BD are chords
- The angle BDA = 39^o
- The angle ABD = 104^o
- The angle BCD = \theta

** Use other angle facts to determine the angle at the centre or the angle at the circumference**.

If we can calculate the size of BAD , we can then use the angle at the centre theorem to calculate BCD . As ABD is a triangle, we can calculate the missing angle BAD as **angles in a triangle total** = 180^o .

BAD = 18 - (104 + 39)

BAD = 37°

**Use the angle at the centre theorem to state the other missing angle**.

The **angle at the centre is twice the angle at the circumference** and so as we know the angle at the circumference BAC , we need to multiply this number by 2 to get the angle BCD at the centre:

BCD = 37 × 2

BCD = 74°

ABCD is a quadrilateral inscribed in a circle with centre C . EF and CG are tangents that touch the circle at B and D . The tangents intersect at the point P . Calculate the size of angle BAD .

** Locate the key parts of the circle for the theorem**.

Here we have:

- BC = CD = Radii
- AB and AD are chords
- The angle FPG = 156^o
- The angle BAD = \theta

** Use other angle facts to determine the angle at the centre or the angle at the circumference**.

If we can calculate the size of BCD , we can then use the angle at the centre theorem to calculate BAD . As EPF is a straight line, we can calculate the angle BPD and then use the tangent theorem to work out BCD . As **angles on a straight line total** = 180^o .

BPD = 180 - 156

BPD = 24°

The **angle between the tangent and the radius is ** 90^o so angle CBP = angle CDP = 90^o .

BCDP is a kite. **Angles in a quadrilateral total ** 360^o so we can work out BCD :

BCD = 360 - (90 + 90 + 24)

BCD = 66°

We now have the following information added to the diagram:

**Use the angle at the centre theorem to state the other missing angle**.

The **angle at the centre is twice the angle at the circumference** and so as we know the angle at the circumference BCD , we need to multiply this number by 2 to get the angle BAD at the centre:

\begin{aligned} &BAD=66 \div 2 \\\\ &BAD=33^o \end{aligned}

ABC is a triangle inscribed in the circle with centre O . The tangent DE touches the circle at point C . Calculate the size of angle BOC .

** Locate the key parts of the circle for the theorem**.

Here we have:

- OB = OC = Radii
- AB, AC and BC are chords
- The angle BCE = 82^o
- The angle BOC = \theta

** Use other angle facts to determine the angle at the centre or the angle at the circumference**.

The angle BAC is in the alternate segment to the angle BCE and so as **angles in the alternate segment are equal**,** **angle BAC = 82^o .

**Use the angle at the centre theorem to state the other missing angle**.

The **angle at the centre is twice the angle at the circumference** and so as we know the angle at the circumference BAC , we need to multiply this number by 2 to get the angle BOD at the centre:

BOC = 82 × 2

BOC = 164 °

**Add to 90/180/360 degrees**

The angle at the circumference added to the angle at the centre is equal to:

– 90^o if the angle is acute.

– 180^o if the angle at the centre is obtuse.

– 360^o if the angle at the centre is a reflex angle.

**Halving and doubling**

By remembering the theorem incorrectly, the student will double the angle at the centre, or half the angle at the circumference.

Top tip: look at the angles. The angle at the centre is always larger than the angle at the circumference (this isn’t so obvious when the angle at the circumference is in the opposite segment).

**Angles are the same**

The angle at the circumference and the centre is incorrectly assumed to be equal.

1. ABCD is an arrowhead inscribed in a circle. The point B is the centre of the circle. Calculate the size of angle ABC , labelled \theta .

36^o

18^o

144^o

108^o

72 \times 2=144^o (angle at the centre is twice the angle at the circumference)

2. Below is a circle with centre D. A, B, and C lie on the circumference of the circle where AB is the diameter of the circle. Calculate the size of angle ABC .

236^o

59^o

62^o

160^o

118 \div 2=59^o (angle at the centre is twice the angle at the circumference)

3. ABCD is a cyclic quadrilateral. E is the centre of the circle. Calculate the size of the reflex angle at E .

236^o

31^o

124^o

298^o

62\times2=124^o (angle at the centre is twice the angle at the circumference)

360 – 124 = 236^o (angles at a point)

4. A, B, and C are points on the circle with radius CD . Calculate the size of angle BAC .

47^o

94^o

43^o

23.5^o

180-(74+59) = 47^o

47 \div 2=23.5^o (angle at the centre is twice the angle at the circumference)

5. A, B and C are points on the circle with centre D . EF and GH are tangents that meet outside the circle at the point P . Calculate the size of the angle ACD .

96^o

114^o

132^o

264^o

APB = 48^o (vertically opposite)

ADB = 180 – 48 = 132^o

Reflex at ADB = 360 – 132 = 228^o (angles at a point)

228 \div 2=114^o (angle at the centre is twice the angle at the circumference)

6. The points A, B, and C lie on the circumference of a circle with centre O . The line DE is a tangent to the circle at point B . Calculate the size of angle AOB .

36^o

72^o

108^o

144^o

ACB = ABE = 72^o (alternate segment theorem)

72 \times 2=144^o (angle at the centre is twice the angle at the circumference)

1. Calculate the size of angle BCD .

**(4 marks)**

Show answer

Reflex angle at BCD = 123 \times 2=246^o

**(1)**

Angle at the centre is twice the angle at the circumference

**(1)**

Angle BCD = 360-246 = 114^o

**(1)**

Angles at a point total 360^o

**(1)**

2.

(a) A, B, C, D are points on the circumference of a circle with centre O . BD is the diameter of the circle. FG is a tangent at the point B and is parallel to the line AC . Angle ABF = 74^o . Calculate the size of angle BOC .

(b) Hence or otherwise, calculate the size of angle y.

**(8 marks)**

Show answer

(a)

BAC = 74^o

**(1)**

Alternate angles in parallel lines are the same

**(1)**

BOC = x= 74\times2 = 148^o

**(1)**

Angle at the centre is twice the angle at the circumference

**(1)**

(b)

COD = 180-148 = 32^o

**(1)**

Angles on a straight line total 180^o

**(1)**

ODC = COD so y=32^o

**(1)**

ODC is an isosceles triangle

**(1)**

3. The circle below has centre C. Points A, B, and D lie on the circumference. Calculate the size of angle BCD , marked \theta .

**(2 marks)**

Show answer

BCD = 66 \times 2=132^o

**(1)**

The angle at the centre is twice the angle at the circumference

**(1)**

You have now learned how to:

- Apply and prove the standard circle theorems concerning angles, radii, tangents and chords, and use them to prove related results

- Alternate segment theorem
- Angles in a semicircle
- Angles in the same segment
- Cyclic quadrilaterals
- Tangents of a circle
- Chord of a circle

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