Circle theorems

In a nutshell

Circle geometry (or circle theorems) involves looking at angle properties when lines and shapes are drawn in circles. Once you can identify and label parts of a circle, it is possible to add lines, triangles and quadrilaterals inside a circle. Circle theorems can help find missing angles.

Circle properties

You should be able to label parts of a circle.

NAME	DESCRIPTION	ILLUSTRATION
Radius	A line from the centre to the edge of the circle.
Diameter	A line across the circle, going through the centre.
Circumference	The perimeter of the circle.
Arc	A section of the circumference.
Chord	A line segment joining two points on the circumference.
Sector	Part of a circle bound by an arc and two radii.
Segment	Part of a circle bound by a chord and an arc.

Circle theorems

There are eight circle theorems that you should learn.

THEOREM	DESCRIPTION	ILLUSTRATION
Angles in the same segment are equal.	$\angle CAD = \angle CBD$
The angle at the centre is twice the angle at the circumference.	$\angle COB = 2 \times \angle CAB$
The angle in a semicircle is $90\degree$ .	$\angle ACB = 90\degree$
The radius and tangent are perpendicular.	$\angle CAP = 90\degree$
The perpendicular bisector of a chord goes through the centre.	$AB \ \perp \ OC$
Opposite angles in a cyclic quadrilateral add up to $180\degree$ .	$\angle a + \angle b = 180\degree$ $\angle c+ \angle d = 180\degree$
Two tangents from the same point are the same length.	$AC = BC$
The alternate segment theorem.	The angle at the tangent is equal to the angle in the alternate segment.

Example 1

In the circle shown, $\angle AOD=120\degree$ . Find $\angle ABD$ , $\angle DBC$ and $\angle DOC$ .

Maths; Angles and geometry; KS4 Year 10; Circle theorems - Higher

$\angle AOD$ is twice the angle at the circumference $\angle ABD$ .

$\begin{aligned}\angle AOD &= 2 \times \angle ABD \\\angle ABD &= \dfrac {120}{2}\\\angle ABD &= \underline{60\degree}\end{aligned}$

$\angle ABC = 90\degree$ as it is the angle in a semicircle. Therefore $\angle ABD$ and $\angle DBC$ add up to $90\degree$ .

$\angle DBC = 90\degree - 60\degree =\underline{30\degree}$

$AOC$ is a straight line, so $\angle DOC$ can be found.

$\begin{aligned}\angle DOC + 120\degree &= 180\degree \\\angle DOC &= 180\degree - 120\degree \\\angle DOC &= \underline{60\degree}\end{aligned}$