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Parts of a circle Angles in polygons Angles on a straight line Angles around a point Angles in parallel lines Triangles Congruence and similarityThis topic is relevant for:

Here we will learn about circle theorems, including their application, proof, and how to use them 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.

Circle theorems are properties that show relationships between angles within the geometry of a circle. We can use these theorems along with prior knowledge of other angle properties to calculate missing angles, without the use of a protractor. This has very useful applications within design and engineering.

There are seven main circle theorems:

- Alternate segment circle theorem
- Angle at the centre circle theorem
- Angles in the same segment circle theorem
- Angle in a semi circle theorem
- Chord circle theorem
- Tangent circle theorem
- Cyclic quadrilateral circle theorem

Below is a summary of each circle theorem, along with a diagram. You need to remember all of these circle theorem rules and be able to describe each one in a sentence.

Download our free circle theorems poster to focus your revision!

The angle that lies between a tangent and a chord is equal to the angle subtended by the same chord in the alternate segment.

**Step-by-step guide:** Alternate segment theorem

To use the alternate segment theorem

- Locate the key parts of the circle for the theorem.
- Use other angle facts to determine one of the two angles.
- Use the alternate segment theorem to state the other missing angle.

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

**Step-by-step guide:** Angle at the centre is twice the angle at the circumference

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.

Angles in the same segment are equal.

**Step-by-step guide:** Angles in the same segment are equal

To use the fact that angles in the same segment are equal

- Locate the key parts of the circle for the theorem.
- Use other angle facts to determine an angle at the circumference in the same segment.
- Use the angle in the same segment theorem to state the other missing angle.

The angle in a semicircle is 90 degrees.

**Step-by-step guide:** Angle in a semicircle

To use the fact that angles in a semicircle equal 90^o

- Locate the key parts of the circle for the theorem.
- Use other angle facts to determine angles within the triangle.
- Use the angles in a semicircle theorem to state the other missing angle.

The perpendicular from the centre of a circle to a chord bisects the chord (splits the chord into two equal parts).

**Step-by-step guide:** Chord of a circle

To find missing angles and lengths using chords

- Locate the key parts of the circle for an appropriate circle theorem.
- Use other angle facts to determine any missing angles.
- Use Pythagoras’ theorem or trigonometry to find the missing length.

The angle between a tangent and radius is 90 degrees. Tangents which meet at the same point are equal in length.

Diagram 1 Diagram 2

**Step-by-step guide:** Tangent of a circle

To use the tangent of a circle:

- Locate the key parts of the circle for the theorem.
- Use other angle facts to determine the remaining angle(s) made with the tangent.
- Use the tangent theorem to state the other missing angle.

The opposite angles in a cyclic quadrilateral total 180^o **.**

**Step-by-step guide:** Cyclic quadrilateral

To use the cyclic quadrilateral theorem

- Locate the key parts of the circle for the theorem.
- Use other angle facts to determine one of the two opposing angles in the quadrilateral.
- Use the cyclic quadrilateral theorem to state the other missing angle.

An angle within a circle is created by two chords meeting at a point on the circumference. The diagrams below show the angle subtended by arc AC from point B for two different circles.

Top tip: The word subtend is used a lot within circle theorems so make sure you know what it means.

**Step-by-step guide:** Subtended angles

Get your free circle theorems worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEGet your free circle theorems worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEThe triangle ABC is inscribed in a circle with centre O. The tangent DE meets the circle at the point A. Calculate the size of the angle ABC.

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

Here we have:

- The tangent DE
- The chord AC (that meets the tangent)
- The angle CAE = 56^o
- The angle ABC = \theta

2**Use other angle facts to determine one of the two angles**

We already know that CAE = 56^o so we do not need to use any other angle fact to determine this angle for this example.

3**Use the alternate segment theorem to state the other missing angle**

ABC = 56^o as angles in the alternate segment are equal to the angle between the tangent and the associated chord.

ABCD 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

**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.

**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:

\begin{aligned} &BAD=150 \div 2 \\\\ &BAD=75^o \end{aligned}

Below is a circle with centre O. AC and BD are chords. Calculate the size of angle CAD.

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

Here we have:

- The angle CBD = 47^o
- AC and BD are chords
- The angle CAD = \theta

**Use other angle facts to determine an angle at the circumference in the same segment**

We already know that CBD = 47^o so we do not need to use any other angle fact to determine this angle for this example.

**Use the angle in the same segment theorem to state the other missing angle**

The angle CAD is in the same segment as the angle CBD and so we can state the angle of CAD

CAD=CBD=47^o

ABC is a triangle where A, B, and C lie on the circumference and AB is a diameter. Angle ABC is 67^o . Calculate the size of angle BAC.

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

Here we have:

- The angle ABC = 67^o
- AB is a diameter
- AC and BC are chords
- The angle CAB = \theta

**Use other angle facts to determine angles within the triangle**

We already know that ABC = 67^o so we do not need to use any other angle fact to determine this angle for this example.

**Use the angles in a semicircle theorem to state the other missing angle**

As the angle in a semicircle is equal to 90^o , angle ACB = 90^o . We can therefore use the fact that angles in a triangle total 180^o to calculate the size of angle BAC:

\begin{aligned}
&BAC=180-(90+67)\\\\
&BAC=23^o
\end{aligned}

Below is a circle with centre C. Points A, B, C, and D are on the circumference of the circle. The chord AB is perpendicular to the line CD at the point E. The line AE is 5cm and angle ADE = 71^o . Calculate the length of the line BC correct to 1 decimal place.

**Locate the key parts of the circle for an appropriate circle theorem**

Here we have:

- CD is a diameter
- AB is a chord, perpendicular to CD
- The angle ADE = 71^o
- The angle BEC = 90^o Radii
- The line AE = 5cm
- The line BC = x

**Use other angle facts to determine other necessary angles**

As angles in the same segment are equal, angle ADE is equal to angle ABC so angle ABC = 71^o . Also, as the perpendicular from the centre of a circle to a chord bisects the chord, the line BE is equal to AE, so BE = 5cm .

**Use Pythagoras’ theorem or trigonometry to find the missing length**

To calculate the length of BC, we need to use trigonometry as we know one side length and two angles where one angle is 90^o .

As we know the side adjacent to the angle and we want to calculate the hypotenuse, we need to use \cos(\theta)=\frac{A}{H} with H as the subject.

\begin{aligned}
&H=\frac{A}{\cos(\theta)}\\\\
&x=\frac{5}{\cos(71)}\\\\
&x=15.4cm \; (1dp)
\end{aligned}

Points A, B, and C are on the circumference of a circle with centre O. DE is a tangent at point A. Calculate the size of angle BAD.

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

Here we have:

- The angle BCA = 52^o
- AC is a diameter
- DE is a tangent
- The angle BAD = \theta

**Use other angle facts to determine the remaining angle(s) made with the tangent**

As AC is a diameter and the angle in a semicircle is 90^o , angle ABC = 90^o . As angles in a triangle total 180^o ,

\begin{aligned} &CAB=180-(90+52)\\\\ &CAB=38^o \end{aligned}

**Use the tangent theorem to state the other missing angle**

As **the angle between the tangent and the radius is ** 90^o , we can now calculate angle BAD:

\begin{aligned} &BAD=90-38 \\\\ &BAD=52^o \end{aligned}

Note: This angle could have also been found using the Alternate Segment Theorem.

ABCD is a quadrilateral where A, B, C and D lie on the circumference of a circle. Calculate the size of angle BCD.

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

Here we have:

- The angle BAD = 51^o
- The angle BCD = \theta

**Use other angle facts to determine one of the two opposing angles in the quadrilateral**

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

**Use the cyclic quadrilateral theorem to state the other missing angle**

As opposite angles in a cyclic quadrilateral total 180^o , we can calculate the size of angle BCD:

\begin{aligned}
&BCD=180-51\\\\
&BCD=129^o
\end{aligned}

Below are some of the common misconceptions for all of the circle theorems:

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

Make sure you know the other angle facts including:

Angles on a straight line total 180^o

Angles at a point total 360^o

Angles in a triangle total 180^o

Angles in a quadrilateral total 360^o

**Halving and doubling**

By remembering the angle at the centre 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**

Make sure that you know when two angles are equal. Look out for isosceles triangles and the angles in the same segment.

**Cyclic quadrilateral**

The angles that are either end of the diameter total 180^o as if the triangle were a cyclic quadrilateral. They should total 90^o as the angle in a semicircle is 90^o .

**A diameter or a chord?**

The angle at the circumference is assumed to be 90^o when the associated chord does not intersect the centre of the circle and so the diagram does not show a semicircle.

**Parallel lines (alternate segment theorem)**

Take for example the diagram below:

The chord BC is assumed to be parallel to the tangent and so the angle ABC is equal to the angle at the tangent. Here the angle BCA would be equal.

Top tip: Use arrows to visualise which way the alternate segment angle appears:

**Opposite angles in a cyclic quadrilateral**

The angle is taken from 180^o which is a confusion with opposite angles in a cyclic quadrilateral.

Here, angle ABC is incorrectly calculated as 180 - 56 = 124^o .

The angle ABC = 56^o as it is in the alternate segment to the angle CAE.

**Angle at the centre is supplementary to opposing angle**

As the shape is a quadrilateral, the angle at the centre is assumed to be supplementary and add to 180^o .

**Opposite angles are the same for a cyclic quadrilateral**

As angles in the same segment are equal, the opposing angles in a quadrilateral are assumed to be equal. The only case of this is when both angles are 90^o .

**Pythagoras’ theorem missing side**

The missing side is calculated by incorrectly adding the square of the hypotenuse and a shorter side, **or** subtracting the square of the shorter sides.

**Incorrect trigonometric function**

The incorrect trigonometric function is used and so the side or angle being calculated is incorrect. This also includes the inverse trigonometric functions.

**Incorrect assumption of isosceles triangles**

The intersection of the diameter and the chord at 90 degrees can be very close to the centre and so the two lengths coming from the point of intersection to the radius are assumed to be equal, but they aren’t.

1. A, B, and C are points on the circumference of a circle with centre O. The tangent DE passes through the point A. Calculate the size of angle ABC.

8^o

82^o

90^o

98^o

ABC = CAE = 82^o (alternate segment theorem)

2. 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

108^o

144^o

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

3. A circle with centre O has four points on the circumference, A, B, C, and D. Angle CAD = 11^o . Calculate the size of the angle CBD.

11^o

79^o

22^o

10^o

CBD = CAD = 11^o (angles in the same segment are equal)

4. The points A, B, and C are on the circumference of a circle. Calculate the size of angle BCA.

127^o

37^o

53^o

90^o

ABC = 90^o (angles in a semicircle)

BCA = 180 \; – \; (90+53) = 37^o

5. A, B, C and D are points on a circle with centre O. BD is the diameter and AC is a chord that is perpendicular to the diameter at E. BE = 2cm and CDE = 30^o . Calculate the length of x , the distance between C and E.

4cm

\frac{\sqrt(3)}{6}cm

2\sqrt{3}cm

2cm

AE = x

BAE = 30^o (angles in the same segment are equal)

x=\frac{2}{\tan(30)}=2\sqrt{3}

6. A circle with centre O has three points on the circumference, A, B, and C. The line DE is a tangent to the circle at the point A. AC is a diameter and angle BAD = 78^o . Calculate the size of the angle ACB.

12^o

78^o

90^o

68^o

BAC = 90 \; – \; 78 = 12^o (the tangent meets the radius at 90^o )

ABC = 90^o (angles in a semicircle)

ACB = 180 \; – \;(90+12) = 78^o (angles in a triangle)

7. ABCD is a cyclic quadrilateral. Calculate the size of angle ABC.

63^o

126^o

127^o

117^o

180-63=117^o (opposite angles in a cyclic quadrilateral total 180^o .)

1. Triangle ABC is inscribed in a circle with centre O. DE is a tangent to the circle at A and FC is a straight line through B. Calculate the size of angle FBA. Explain your working.

**(2 marks)**

Show answer

Angle ABC = 72^o as it is in the alternate segment to angle CAE.

**(1)**

FBA = 180-72 = 108^o as angles on a straight line total 180^o .

**(1)**

2. A, B, C, and D are points on the circumference of a circle. AE = BE. Show that the point E is **not **the centre of the circle. Explain any assumptions made.

**(7 marks)**

Show answer

DAE = 51^o as angles in the same segment are equal.

**(1)**

AED = 180-(51+47) = 82^o as angles in a triangle total 180^o .

**(1)**

AEB = 180-82 = 98^o as angles on a straight line total 180^o .

**(1)**

EAB = EBA = (180-98)\div 2 = 41^o

**(1)**

ABC = 51 + 41 = 92^o

**(1)**

If E was the centre then angle ABC = 90^o as the angle in a semicircle is 90^o .

**(1)**

E is not the centre as ABC = 92^o .

**(1)**

3. Triangle ABC is inscribed in a circle with centre O.

Angle AOB = 120^o and angle ACO = 30^o . Show that the triangle is equilateral. Explain any assumptions made in your working.

**(7 marks)**

Show answer

OA = OB = OC as all radii.

**(1)**

Triangles AOB, AOC and BOC are all isosceles.

**(1)**

OAC = 30^o and AOC = 120^o

**(1)**

OAB = OAC = 30^o

**(1)**

BOC = 360-240 = 120^o and angles at a point total 360^o .

**(1)**

OBC = OCB = 30^o

**(1)**

Angles ABC = BCA = CAB = 60^o and triangle is equilateral.

**(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

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