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Properties of Circles

Every measurement on a circle grows from one number: the radius. Name the parts, learn two formulas for the whole circle, then scale them down by a fraction of the turn to handle arcs and sectors.

MYP 5Standard MathsCirclesCriteria A · C · D~10 min read

Written by Fayek Jawad Haider · MYP student

Parts of a circle

Getting the names right makes every formula easier to read, because each one is written in terms of these parts.

Radius
The distance from the centre to the edge. Written \(r\), it is the building block of every circle formula.
Diameter
A straight line right across the circle through the centre. It is twice the radius, so \(d = 2r\).
Circumference
The distance all the way around the edge, the circle's own perimeter.
Arc and sector
An arc is part of the circumference; a sector is the pie-slice region between two radii and the arc joining them.

Circumference and area

Two formulas describe the whole circle. Both rely on \(\pi\) (pi), the fixed number, roughly \(3.142\), that links a circle's size to its circumference.

\[ C = 2\pi r \qquad\qquad A = \pi r^2 \]

Circumference \(C\) uses the radius once; area \(A\) uses it squared. If you are given the diameter instead, halve it first to get \(r\).

Worked example

A circle has radius \(7\) cm. Find its circumference and its area, each to 1 decimal place.

1
Circumference: \(C = 2\pi r = 2 \times \pi \times 7 = 14\pi\).
2
Evaluate: \(14\pi = 43.98\ldots\), so \(C \approx 44.0\) cm.
3
Area: \(A = \pi r^2 = \pi \times 7^2 = 49\pi = 153.93\ldots\).
\(C \approx 44.0\) cm and \(A \approx 153.9\) cm\(^2\)

Square the radius, not the pi

In \(A = \pi r^2\) only the radius is squared. Work out \(r^2\) first, then multiply by \(\pi\). A length answer is in cm and an area answer is in cm\(^2\), so always attach the right unit.

Arc length

An arc is just a fraction of the full circumference. The fraction is the angle at the centre, \(\theta\), out of the full \(360^\circ\).

\[ \text{arc length} = \frac{\theta}{360} \times 2\pi r \]

So a \(90^\circ\) arc is a quarter of the circumference, a \(180^\circ\) arc is half, and so on. Take the whole-circle formula and scale it by \(\dfrac{\theta}{360}\).

Worked example

A sector has radius \(10\) cm and a centre angle of \(72^\circ\). Find the length of its arc, to 1 decimal place.

1
Write the formula: arc length \(= \dfrac{\theta}{360} \times 2\pi r\).
2
Substitute: \(= \dfrac{72}{360} \times 2\pi \times 10 = 0.2 \times 20\pi\).
3
Evaluate: \(0.2 \times 62.83\ldots = 12.56\ldots\).
Arc length \(\approx 12.6\) cm

Sector area

A sector area follows exactly the same idea: take the same fraction of the whole circle's area.

\[ \text{sector area} = \frac{\theta}{360} \times \pi r^2 \]

The only change from arc length is which whole-circle formula you scale: use \(2\pi r\) for the curved edge, and \(\pi r^2\) for the region inside.

Worked example

Find the area of the same sector: radius \(10\) cm and centre angle \(72^\circ\), to 1 decimal place.

1
Write the formula: sector area \(= \dfrac{\theta}{360} \times \pi r^2\).
2
Substitute: \(= \dfrac{72}{360} \times \pi \times 10^2 = 0.2 \times 100\pi\).
3
Evaluate: \(0.2 \times 314.15\ldots = 62.83\ldots\).
Sector area \(\approx 62.8\) cm\(^2\)

Where this is assessed

Recalling and using the right formula is Criterion A (Knowing and understanding). Showing the fraction \(\dfrac{\theta}{360}\), the substitution and the units is Criterion C (Communicating). Applying it to pizza slices, fan-shaped windows and running tracks is Criterion D (Applying mathematics in real-life contexts). Check the units match the quantity: arc length in cm, sector area in cm\(^2\).

Check yourself

1. A circle has radius \(5\) cm. Find its area, to 1 decimal place. +

Use \(A = \pi r^2 = \pi \times 5^2 = 25\pi = 78.53\ldots\), so \(A \approx \mathbf{78.5}\) cm\(^2\).

2. A circle has diameter \(20\) cm. Find its circumference, to 1 decimal place. +

Halve the diameter to get \(r = 10\) cm, then \(C = 2\pi r = 2\pi \times 10 = 20\pi = 62.83\ldots\), so \(C \approx \mathbf{62.8}\) cm. (Equivalently, \(C = \pi d = 20\pi\).)

3. A sector has radius \(8\) cm and a centre angle of \(90^\circ\). Find its arc length, to 1 decimal place. +

Arc length \(= \dfrac{90}{360} \times 2\pi \times 8 = 0.25 \times 16\pi = 4\pi = 12.56\ldots\), so the arc \(\approx \mathbf{12.6}\) cm.

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