Circle Theroms Maths Questions | Worksheets and Revision | MME

Circle Theorems Questions, Worksheets and Revision

Level 8-9
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The 8 Circle Theorems.

You should be familiar with all 8 circle theorems to the point where:

a) You can identify when they should be used.

b) You can describe which one you’ve used with appropriate language.

Make sure you are happy with the following topics before continuing 

Note: Circle geometry problems often require knowledge of all the basic geometry rules in order to solve them. 

Level 6-7

Rule 1

Angles in the same segment are equal.

\textcolor{red}{x} = \textcolor{red}{x} 

Triangles drawn from the same chord will have the same angle when touching the circumference. 

Level 6-7

Rule 2

Opposite angles in a cyclic quadrilateral add up to 180\degree.

\textcolor{purple}{w}+\textcolor{red}{x}=180\degree

\textcolor{blue}{y}+\textcolor{limegreen}{z}=180\degree

This is a 4 sided shape with every corner touching the circumference of the circle. 

Level 6-7

Rule 3

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

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

Level 6-7

Rule 4

The perpendicular bisector of a chord passes through the centre of the circle

A line perpendicular and in the centre of a chord (a line drawn across the circle) will always pass through the centre of the circle. 

Level 6-7
Level 6-7

Rule 5

The radius will always meet a tangent to the circle at 90\degree.

A tangent (a line touching a single point on the circumference) will always make an angle of exactly 90\degree with the radius.

You can say that a tangent and radius that meet are perpendicular to each other. 

Level 6-7

Rule 6

The tangents from the same point to a circle are equal in length.

AB = BC

Two tangents (a line touching a single point on the circumference) drawn from the same outside point are always equal in length. 

Level 6-7

Rule 7

The angle inscribed in a semicircle is always a right angle. 

A triangle drawn with the diameter will always make a 90\degree angle where it hits the circumference.

Another way of saying this is that a diameter ‘subtends’ a right-angle at the circumference.

Level 6-7

Rule 8

Alternate Segment Theorem: The angle between the tangent and the side of the triangle is equal to the opposite interior angle.

\textcolor{limegreen}{x}=\textcolor{limegreen}{x}

\textcolor{red}{y}=\textcolor{red}{y}

The angle between the tangent and the triangle will be equal to the angle in the alternate segment.

(This is the hardest rule and can be tricky to spot).

Level 6-7
Level 6-7

Example 1: Angle in a Semi-Circle

Below is a circle with centre C.

BD is a diameter of the circle, A is a point on the circumference.

What is the size of angle CBA?

[2 marks]

If a question says “show our workings”, you must state what circle theorem/geometry fact you use when you use it.

BD is a diameter of the circle, we know that triangle BAD is confined within the semi-circle.

So we can use Rule 7, the angle in a semi-circle is a right-angle to deduce that \angle BAD = 90\degree.

To find CBA, we just need to subtract from 180\degree.

\angle CBA = 180\degree - 23\degree - 90\degree = 67\degree

Level 6-7

Example 2: Combining Rules

Below is a circle with centre C.

A, B, and D are points on the circumference.

Angle \angle BCD is 126\degree and angle \angle CDA is 33\degree.

Find angle ABC.

You must show your workings.

[2 marks]

The first circle theorem we’re going to use here is: Rule 3, the angle at the centre is twice the angle at the circumference.

The angle at the centre is 126\degree, so;

\angle BAD = 126\degree \div 2 = 63\degree.

We now know two out of the four angles inside ABCD.

To find a third, simply observe that angles around a point sum to \textcolor{orange}{360\degree}:

360\degree - 126\degree = 234\degree

Since the angles in a quadrilateral sum to \textcolor{orange}{360\degree}, we can find the angle we’re looking for.

\angle ABC= 360\degree - 33\degree - 63\degree - 234\degree = 30 \degree

Level 8-9
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Example Questions

The angle at the centre is twice the angle at the circumference. Given that the angle formed at the centre, which in this case is 98\degree, is exactly twice the angle at the circumference of a circle at the same point. We simply have to divide the angle at the centre of the circle by two: 

 

x=98\degree \div 2 = 49\degree

The angle in a semicircle is always a right angle. Given that any triangle drawn with the diameter will always make a 90° angle where it hits the opposite circumference. We can also use that interior angles in a triangle add up to 180°, we find that, 

 

x=180\degree - 90\degree - 32\degree = 58\degree

Firstly, recognise that since BD is a diameter, angle BAD is the angle in a semi-circle. Our circle theorems tell us that the angle in a semi-circle is a right-angle so BAD must be 90\degree. As we now know this, we get that

 

\text{Angle BAE } = 90 + 31 = 121 \degree.

 

Next, we recognise that ABDE is a cyclic quadrilateral. On a related note, the second circle theorem we’re going to use is: opposite angles in a cyclic quadrilateral sum to 180. Angle BAE (which we just worked out) is opposite to angle CDE, so

 

\text{Angle CDE } = 180 - 121 = 59\degree

 

Then, the final step to finding angle EDA will be subtracting the size of angle CDA from that of angle CDE to get

 

\text{Angle EDA } = 59 - 18 = 41\degree.

Our first circle theorem here will be: tangents to a circle from the same point are equal, which in this case tells us that AB and BD are equal in length.

 

This means that ABD must be an isosceles triangle, and so the two angles at the base must be equal. In this case those two angles are angles BAD and ADB, neither of which know. Let the size of one of these angles be x, then using the fact that angles in a triangle add to 180, we get

 

x + x + 42 = 180\degree.

 

Then, subtract 42 from both sides to get

 

2x = 180 - 42 = 138\degree,

 

and divide both sides by 2 to get

 

x = 69\degree.

 

Now we can use our second circle theorem, this time the alternate segment theorem. This tells us that the angle between the tangent and the side of the triangle is equal to the opposite interior angle. Given that angle ADB, which is 69\degree, is the angle between the side of the triangle and the tangent, then the alternate segment theorem immediately gives us that the opposite interior angle, angle AED (the one we’re looking for), is also 69\degree.

Firstly, using the fact angles inside of a triangle add together to 180\degree. Angle ABC is, 

 

180\degree-71\degree-23\degree=86\degree

 

The Alternate Segment Theorem states the angle between the tangent and the side of the triangle is equal to the opposite interior angle, hence

 

x\degree=\angle \text{ABC} = 86 \degree.

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