Parts of a circle and equal chords

The parts of a circle

Fix a pin in a board, tie a string to it, and pull the string tight with a pencil. Now move the pencil all the way round. The pencil draws a circle. Every point of the circle is the same distance from the pin.

• The pin is the centre. We call it $O$.
• The fixed distance is the radius. All the radii of one circle are equal.
• A line segment that joins two points on the circle is a chord.
• A chord that passes through the centre is a diameter. It is twice as long as the radius.
• A piece of the circle is an arc. Two points on a circle cut it into two arcs. The shorter one is the minor arc and the longer one is the major arc.
• When the two points are the ends of a diameter, the two arcs are equal. Each one is a semicircle.
• The length of the whole circle is its circumference.
• The region between a chord and an arc is a segment. The region between an arc and the two radii at its ends is a sector.

Two circles with the same radius are congruent. Everything in this chapter holds for one circle, and it holds in the same way for two equal circles.

The diameter is the longest chord. Take any chord $AB$ that is not a diameter, and join $A$ and $B$ to the centre $O$. In triangle $OAB$ the side $AB$ is shorter than $OA + OB$, which is two radii. A diameter is exactly two radii long. So no chord is longer than a diameter.

Chords and the angle at the centre

Draw a chord $AB$ and join its ends to the centre $O$. The angle $AOB$ is the angle that the chord subtends at the centre. A short chord subtends a small angle. A long chord subtends a large one. A diameter subtends $180^\circ$.

Do two chords of the same length subtend the same angle? Yes, and here is the proof.

Theorem 1. Equal chords of a circle subtend equal angles at the centre.

Given. A circle with centre $O$, and two chords $AB$ and $CD$ with $AB = CD$.

To prove. $\angle AOB = \angle COD$.

Proof. Take the triangles $AOB$ and $COD$.

1. $OA = OC$ and $OB = OD$, because they are radii of the same circle.
2. $AB = CD$. This is given.
3. So $\triangle AOB \cong \triangle COD$, by the SSS rule. The corners match in the order $A$ with $C$, $O$ with $O$ and $B$ with $D$.
4. Therefore $\angle AOB = \angle COD$, by CPCT.

Theorem 2 (the converse). If two chords of a circle subtend equal angles at the centre, the chords are equal.

Proof. Take the same two triangles.

1. $OA = OC$ and $OB = OD$, because they are radii.
2. $\angle AOB = \angle COD$. This is given, and each angle lies between the two radii.
3. So $\triangle AOB \cong \triangle COD$, by the SAS rule.
4. Therefore $AB = CD$, by CPCT.

The triangle at the centre

Because $OA = OB$, triangle $OAB$ is isosceles. Its two base angles are equal. So once you know the angle at the centre, you know the other two angles:

A worked example. A chord $AB$ of a circle with centre $O$ makes an angle of $80^\circ$ at $O$. Find $\angle OAB$.

1. The triangle. $OA = OB$, so $\angle OAB = \angle OBA$.
2. The angle sum. $\angle OAB + \angle OBA = 180^\circ - 80^\circ = 100^\circ$.
3. Share it. The two angles are equal, so $\angle OAB = 50^\circ$.

A second example, with algebra. The chords $PQ$ and $RS$ of a circle with centre $O$ are equal. $\angle POQ = (3x + 10)^\circ$ and $\angle ROS = (5x - 30)^\circ$. Find $x$ and $\angle POQ$.

1. Equal chords. By Theorem 1, $\angle POQ = \angle ROS$.
2. An equation. $3x + 10 = 5x - 30$, so $2x = 40$ and $x = 20$.
3. The angle. $\angle POQ = 3 \times 20 + 10 = 70^\circ$. Check: $\angle ROS = 5 \times 20 - 30 = 70^\circ$.

A chord as long as the radius. If $AB$ is equal to the radius, then $OA = OB = AB$. Triangle $OAB$ is equilateral, so $\angle AOB = 60^\circ$, and the other two angles are $60^\circ$ as well.

Practise with new numbers

Try it in Python

A computer can find the angle that a chord makes at the centre, from its length and the radius. This program does it for five chords of a circle with radius $5$.

Common mistakes

• Mixing up the radius and the diameter. The diameter is twice the radius.
• Thinking that a longer arc means a longer chord only. The chord is the straight segment. The arc is the curved piece.
• Forgetting that the proof needs the radii. The two triangles have two pairs of equal sides only because the sides are radii.
• Giving the angle at the centre as the base angle. In triangle $OAB$ the angle at $O$ is the apex. The angles at $A$ and $B$ are the equal ones.

Remember these

• All radii of a circle are equal, and a diameter is twice the radius and the longest chord.
• Equal chords subtend equal angles at the centre, and chords that subtend equal angles at the centre are equal.
• In triangle $OAB$, $\angle OAB = \angle OBA = \dfrac{180^\circ - \angle AOB}{2}$.
• A chord equal to the radius subtends $60^\circ$ at the centre.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A circle has a diameter of $18\ \text{cm}$. What is its radius, and what is the length of its longest chord?

2. The chords $AB$ and $CD$ of a circle with centre $O$ are equal, and $\angle AOB = 58^\circ$. What is $\angle COD$?

3. A chord $AB$ makes an angle of $70^\circ$ at the centre $O$. Find $\angle OAB$.

4. A chord $PQ$ is $9\ \text{cm}$ long and the radius of the circle is $9\ \text{cm}$. Find $\angle POQ$.

5. Why is a diameter longer than any other chord?