What you need to know

Circles are very useful and interesting shapes for us mathematicians, and as a result we gave just about every feature of a circle you can dream up its own special name. You should be familiar with all the terms and their definitions, as labelled on the diagram below.

– Centre: The point which lies at the centre of the circle.

– Radius: Any line from the centre to the circumference.

– Diameter: Any line which touches the circumference at two points and passes through the centre.

– Circumference: The perimeter of the circle.

– Chord: Any line which touches two points on the circumference.

– Segment: The area formed by a chord and the circumference.

– Arc: A fraction of the circumference.

– Sector: The area formed by two radii and an arc (also, a fraction of the total area).

– Tangent: A line which touches the circumference at only one point.

Knowing definitions is not all that interesting, and it feels like there are an awful lot, but try not to get too hung up on them. It’s more important that you understand how to do actual maths problems involving them, and if you practise those enough then lots of the definitions will end up sticking in your head whether you want them to or not.

Some of them, such as tangents and chords are seen more often in the section on Circle Theorems (link?).

For this section, we will first consider the following two formulae.

1. Area of a circle = \pi \times \text{ radius } \times \text{ radius } = \pi r^2

2. Circumference of a circle  = \pi \times \text{ diameter } = \pi d = 2 \pi r

Where r and d are used to mean ‘radius’ and ‘diameter’ respectively, and \pi (spelt like ‘pi’ and pronounced like ‘pie’) is that mysterious never-ending decimal which begins 3.14159265…

Knowing these formulae, you may be given the radius/diameter of a circle and be asked to find the area/circumference, or you may be given the area/circumference and be asked to work backwards to find the radius/diameter.

Example: Below is a circle with centre C. Find its area and circumference to 1 decimal place.

Because the line passes through the centre, we know it is a diameter of the circle. So,

Circumference = \pi d = \pi \times 42 = 131.9 \text{cm (1d.p.)}

The diameter is twice the length of the radius, so the \text{radius } = 42 \div 2 = 21cm. So:

Area of circle = \pi r^2=\pi\times (21^2) = 1385.4\text{cm}^2\text{ (1d.p.)}

Note: some questions will ask you to “leave your answer in terms of pi”, this means that the answer should be in the form of “something \times \pi”. For example, if we were to leave the circumference of this circle in terms of pi, it would simply be written like 42\pi.

Sectors & Arc Lengths

In addition to working out areas and circumferences, you must be able to solve problems around sectors and arc lengths. This involves thinking of them as fractions of the total area and total circumference respectively, because, well, that’s what they are.

The key point when calculating these values will be to look at what the size of the angle inside the sector is (we’re talking the angle between the two radii and yes, that is the plural of radius). The size of the angle inside the sector as a fraction of 360 (because 360 is the total number of degrees in a circle) will tell us precisely what proportion of the full circle that given sector makes up. Then, all we need to do is multiply this fraction by what the total area (or circumference) of the circle would be (if all of it were drawn out), and voilà, we’ve calculated the area of the sector (or the arc length).

Example: Below is a sector of a circle with centre C. The angle formed by the two radii is 102 degrees and the radius is 8cm. Work out the area of the sector and the arc length to 3 significant figures.

The angle is 102 degrees, which means that this sector constitutes a fraction of \frac{102}{360} of the whole circle. So, we get:

\text{Sector area }= \dfrac{102}{360} \times \pi \times 8^2

= 57.0\text{cm}^2

\text{Arc length } = \dfrac{102}{360} \times \pi \times 2 \times 8

= 14.2\text{cm}

Example Questions

The formula for the area of a circle is \pi r^2. In this question we aren’t given the radius, but we can the line labelled as 5.24cm passes through centre so is a diameter. Given that the diameter is twice the length of the radius, we can work out the length of the radius.

 

\text{radius } = 5.24 \div 2 = 2.62\text{cm}

 

Therefore, \text{area } = \pi r^2 = \pi \times (2.62)^2 = 21.6\text{cm}^2

The question asks for the total perimeter of the shape. We know that one side is 14mm but the other two are missing. Immediately we can identify that the other straight side is also a radius of the circle and so will also be 14mm long. Then, all that remains is to calculate the arc length and add up our answers.

 

The angle in this sector is 165 degrees, meaning that the arc length will be equal to \frac{165}{360} of the total circumference. The formula for the circumference is \pi d, or alternatively (and more helpfully in this case), 2\pi r So, we get:

 

Arc length = \dfrac{165}{360} \times 2 \pi \times 14 = 40.3\text{mm}

 

Therefore, \text{total perimeter } = 14 + 14 + 40.3 = 68.3\text{mm}

Because we know the area this time but not the angle, we’re going to have to set up an equation involving x and then solve it.

 

The formula for the area of a circle is \pi r^2. The area of this sector, 160\text{m}^2, must be equal to \frac{x}{360} of the total area of the circle. So, as an equation, this looks like:

 

160 = \dfrac{x}{360} \times \pi \times 9^2 = \dfrac{x}{360} \times 81\pi

 

Dividing the left-hand side and right-hand side by 81 \pi gives

 

\dfrac{160}{81\pi} = dfrac{x}{360}

 

Then, multiply both sides by 360 to get

 

\dfrac{160}{81\pi} \times 360 = x, so

 

x = 226\degree (to 3s.f.)

Whether you are a GCSE Maths tutor in Leeds or you run a tuition agency in York, this dedicated Circles resources page will come in handy. You may be looking for segment revision worksheets or arc length questions, well you will find it all hear. All worksheets have been manually checked by an expert Maths tutor to ensure they are suitable for the new 9-1 Maths course and relevant to all major exam boards including AQA, OCR and Edexcel.

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