Circles, Arcs and Sectors

The Circle

A circle has three main components:

  • The circumference, which is the outside edge of the circle.
  • The radius, which goes from any point on the circumference to the centre of the circle.
  • The diameter, which joins the circumference through the centre of the circle.
Circle with radius and diameter

Calculating the Circumference

A relationship exists between all circles: the circumference divided by the diameter is always the same.

This ratio is called pi, the 16th letter of the Greek alphabet. It is an irrational number and has the symbol $\pi$.

By definition:

$$ \frac{C}{d} = \pi $$

Re‑arranging gives the formula:

$$ C = \pi d $$

Very often, $\pi$ is taken to 2 decimal places and we use the value $3.14$.

Example

Find the circumference of a circle which has a diameter of $4\text{ cm}$.

Use the $\pi$ button on your calculator and give your answer correct to two decimal places.

$$ C = \pi d = \pi \times 4 \approx 12.57\text{ cm} $$

Since the diameter of a circle is twice its radius, $d = 2r$.

Substituting into $C = \pi d$ gives:

$$ C = \pi (2r) = 2\pi r $$

Example

Find the circumference of a circle which has a radius of $4\text{ cm}$.

Use the $\pi$ button on your calculator and give your answer correct to two decimal places.

$$ C = 2\pi r = 2\pi \times 4 \approx 25.13\text{ cm} $$

Area of a Circle

To find the area of a circle, you could attempt to count the number of squares inside it.

Circle on a grid of squares

This would give an approximation of the area.

You could also cut the circle into segments and lay them out next to each other.

Here, the circle is cut into 8 equal parts.

Circle cut into sectors
Circle sectors rearranged into a rectangle

As the circle is cut into smaller and smaller parts, a rectangle is formed.

Using the equation $\text{Area} = \text{length} \times \text{breadth}$, we obtain:

$$ A = \pi r^2 $$

Example

Find the area of a circle which has a radius of $4\text{ cm}$.

Use the $\pi$ button on your calculator and give your answer correct to two decimal places.

$$ A = \pi r^2 = \pi \times 4^2 = 16\pi \approx 50.27\text{ cm}^2 $$

Example

Find the area of a circle which has a diameter of $4\text{ cm}$.

Use the $\pi$ button on your calculator and give your answer correct to two decimal places.

Since $d = 4\text{ cm}$, the radius is $r = 2\text{ cm}$.

$$ A = \pi r^2 = \pi \times 2^2 = 4\pi \approx 12.57\text{ cm}^2 $$

Interactive Circle Explorer

Use the slider to change the radius. The circle and calculations update instantly.

Radius:

Diameter:

Circumference:

Area:

Arcs and Sectors – Terminology

An arc is a part of a curve. It is a fraction of the circumference of the circle.

Arc of a circle

A sector is part of a circle enclosed between two radii.

Sector of a circle

A chord is a line joining two points on a curve.

Chord in a circle

A chord can be a diameter.

Diameter as a chord

Arcs and Sectors – Equations

If a sector has angle $\theta^\circ$ at the centre, then:

$$ \frac{\theta}{360} = \frac{\text{arc length}}{2\pi r} = \frac{\text{sector area}}{\pi r^2} $$

So:

$$ \text{Arc length} = \frac{\theta}{360} \times 2\pi r $$

$$ \text{Sector area} = \frac{\theta}{360} \times \pi r^2 $$

Arcs

Example

What is the length of arc $AB$?

Arc AB diagram

Use $\text{Arc length} = \dfrac{\theta}{360} \times 2\pi r$.

$$ \begin{aligned} \text{Arc length} &= \frac{\theta}{360} \times 2\pi r \\[6pt] &= \frac{120}{360} \times 2\pi \times 4 \\[6pt] &= \frac{1}{3} \times 8\pi \\[6pt] &= \frac{8\pi}{3} \\[6pt] &= 8.3775\ \text{cm} \\[6pt] &= 8.4\ \text{cm (1 dp)} \end{aligned} $$

Length of arc AB.

Example

Find the radius of the following circle:

Arc length given, find radius
$$ \frac{\text{angle at centre}}{360^\circ} = \frac{\text{length of arc}}{\pi d} $$ $$ \text{angle at centre} \times \pi d = \text{length of arc} \times 360^\circ $$ $$ \pi d = \frac{\text{length of arc} \times 360^\circ}{\text{angle at centre}} $$ $$ d = \frac{\text{length of arc} \times 360^\circ}{\text{angle at centre} \times \pi} $$

Finding the radius from arc length and angle.

$$ d = \frac{ \text{length of arc} \times 360^\circ }{ \text{angle at centre} \times \pi } $$ $$ = \frac{ 3 \times 360^\circ }{ 60 \times \pi } $$ $$ = 5.7295\ \text{cm} $$ $$ \text{but } d = 2r $$ $$ \text{so } r = \frac{5.7295}{2} $$ $$ = 2.8647\ \text{cm} $$ $$ = 2.9\ \text{cm (1 dp)} $$

Sectors

Example

What is the area of sector $AOB$?

Sector AOB diagram

Use $\text{Sector area} = \dfrac{\theta}{360} \times \pi r^2$.

$$ \text{area of sector} = \frac{120^\circ}{360^\circ} \times \pi \times 4^2 $$ $$ \text{area of sector} = \frac{1}{3} \times \pi \times 16 $$ $$ \text{area of sector} = 16.755 $$ $$ \text{area of sector} = 16.8\ \text{cm}^2\ (1\ \text{dp}) $$
Example

Find the radius of the following circle:

Sector area given, find radius

Rearrange $\text{Sector area} = \dfrac{\theta}{360} \times \pi r^2$ to solve for $r$.

$$ \frac{\text{angle at centre}}{360^\circ} = \frac{\text{area of sector}}{\pi r^2} $$ $$ \theta \times \pi r^2 = \text{area of sector} \times 360^\circ $$ $$ \pi r^2 = \frac{\text{area of sector} \times 360^\circ} {\theta} $$ $$ r^2 = \frac{\text{area of sector} \times 360^\circ} {\theta \times \pi} $$ $$ r = \sqrt{ \frac{ \text{area of sector} \times 360^\circ }{ \theta \times \pi } } $$

Substituting to find solution

$$ r = \sqrt{ \frac{ 16 \times 360^\circ }{ 120 ^\circ\times \pi } } $$ $$ r = \sqrt{15.27887} $$ $$ r = 3.90882 $$ $$ r = 3.9\ \text{cm (1 dp)} $$
Example

What is the length of arc $AB$?

Arc AB in a sector

Relating arc length to sector area.

$$ \frac{\text{length of arc}}{\pi d} = \frac{\text{area of sector}}{\pi r^2} $$ $$ \text{length of arc} \times \pi r^2 = \text{area of sector} \times \pi d $$ $$ \text{length of arc} = \frac{ \text{area of sector} \times \pi d }{ \pi r^2 } $$
$$ \text{length of arc} = \frac{16 \times \pi \times 10}{\pi \times 5^2} $$ $$ \text{length of arc} = \frac{160}{25} = 6.4\ \text{cm} $$

Chords, Bisectors and Tangents

Chord and bisector
Chord properties
Chord and centre relationships

A tangent touches the circle at one point only.

Tangent to a circle
Tangent and radius at right angles

Pythagoras in a Circle

Drag the point on the circle. The right triangle updates and you can see $a^2 + b^2 = c^2$ and $x^2 + y^2 = r^2$ in action.

Coordinates of point: (x, y) = (, )

Radius: r =

Pythagoras in the triangle:
$a = |x|,\; b = |y|,\; c = r$
$a^2 + b^2 = $
$c^2 = r^2 = $

Because the point lies on the circle, $x^2 + y^2 = r^2$ and that’s exactly $a^2 + b^2 = c^2$.

Example

What is the value of $x$?

Right triangle in a circle

Move the radius down:

Rearranged triangle

By the theorem of Pythagoras:

$$ a^2 + b^2 = c^2 $$

Using Pythagoras’ theorem to find the missing side.

$$ \text{hypotenuse}^2 = \text{short side}^2 + \text{other side}^2 $$ $$ 6^2 = 5^2 + x^2 $$ $$ 6^2 - 5^2 = x^2 $$ $$ 36 - 25 = x^2 $$ $$ x^2 = 11 $$ $$ x = \sqrt{11} $$ $$ x = 3.3166\ \text{m} $$ $$ x = 3.3\ \text{m (1 dp)} $$

The Angle in a Semi‑Circle

A famous result in geometry says that any angle drawn in a semi‑circle is a right angle. This is sometimes called Thales’ Theorem.

Angle in a semi‑circle = $90^\circ$

Angle: 0°