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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:
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
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
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
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
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
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:
The opposite angles in a cyclic quadrilateral total 180^o .
Step-by-step guide: Cyclic quadrilateral
To use the cyclic quadrilateral theorem
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.
Here we have:
2Use 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.
3Use 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:
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:
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:
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:
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:
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:
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:
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
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).
Make sure that you know when two angles are equal. Look out for isosceles triangles and the angles in the same segment.
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 .
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.
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:
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.
As the shape is a quadrilateral, the angle at the centre is assumed to be supplementary and add to 180^o .
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 .
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.
The incorrect trigonometric function is used and so the side or angle being calculated is incorrect. This also includes the inverse trigonometric functions.
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.
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 .
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.
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.
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.
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.
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.
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)
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)
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)
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:
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