## What you need to know

## The **8 circle theorems.**

You should be familiar with them all 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.

We’ll have a look at an example below but first, here’s all 8 of them.

Make sure you are happy with the following topics before continuing

– Circles

## Rule 1

Angles in the **same segment** are **equal**.

x = x

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

## Rule 2

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

w+x=180\degree

y+z=180\degree

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

## 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 at the same point.

## 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.

## Rule 5

The **radius** is **perpendicular** to any **tangent to the circle**.

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

## 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.

## Rule 7

The **angle 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 opposite circumference.

## Rule 8

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

x=x

y=y

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

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

**Example 1**

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 mark]**

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 \text{angle BAD } = 90\degree.

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

\text{Angle CBA } = 180 - 23 - 90 = 67\degree

**Example 2**

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]**

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

We need to solve this in two steps.

First: 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, so;

\text{angle BAD } = 126 \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 360**, t

360 - 126 = 234\degree

Since the **angles in a quadrilateral sum to 360**, if we subtract the ones we know from 360 then find the angle we’re looking for.

\text{Angle ABC } = 360 - 33 - 63 - 234 = 30 \degree.

### Example Questions

**Question 1: **Points A, B, and C all lie on the circumference of a circle with centre O.

Work out the value of angle x. Give a reason for each step of your answer.

**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

**Question 2: **Points A, B, and C all lie on the circumference of a circle and the line BC passes through the centre of the circle, O.

Work out the value of angle x. Give a reason for each step of your answer.

**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

**Question 3:** Below is a circle with centre C. A, B, D, and E are points on the circumference. BD is a diameter of the circle.

Angle CDA is 18\degree Find the size of angle EDA. You must show your workings.

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.

**Question 4:** Below is a circle with centre C. A, E, and D are points on the circumference. AB and BD are tangents to the circle.

Angle ABD is 42\degree. Find the size of angle AED. You must show your workings.

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.

Then, subtract 42 from both sides to get

2x = 180 - 42 = 138,

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.

**Question 5:** Below is a circle with centre O. A, B, and C are points on the circumference. A tangent to the circles passes through point A.

Given that angle BAC is 23\degree and angle ACB is 71\degree, find the size of angle x\degree.

You must show your workings.

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.

### Worksheets and Exam Questions

#### (NEW) Circle Theorems Exam Style Questions - MME

Level 6-7#### Circle Theorems

Level 6-7#### Circle Theorems 2

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