The perpendicular from the centre

The distance of a chord from the centre

A baker cuts a round cake along a straight line, and the cut is a chord. How far is the cut from the centre of the cake? The shortest way from the centre to the cut is the line that meets it at a right angle. Its length is the distance of the chord from the centre.

Theorem 3. The perpendicular from the centre of a circle to a chord bisects the chord.

Given. A circle with centre $O$, a chord $AB$, and $OM$ perpendicular to $AB$ at $M$.

To prove. $AM = MB$.

Proof. Take the triangles $OMA$ and $OMB$.

1. $\angle OMA = \angle OMB = 90^\circ$, because $OM$ is perpendicular to $AB$.
2. $OA = OB$, because they are radii. These are the hypotenuses.
3. $OM = OM$. This side is common.
4. So $\triangle OMA \cong \triangle OMB$, by the RHS rule.
5. Therefore $AM = MB$, by CPCT.

Theorem 4 (the converse). The line from the centre of a circle to the middle point of a chord is perpendicular to the chord.

Proof. Let $M$ be the middle point of the chord $AB$, so $AM = MB$.

1. $OA = OB$, because they are radii.
2. $AM = MB$. This is given.
3. $OM = OM$. This side is common.
4. So $\triangle OMA \cong \triangle OMB$, by the SSS rule, and $\angle OMA = \angle OMB$.
5. These two angles lie on the straight line $AB$, so they add up to $180^\circ$. Each one is $90^\circ$.

Finding the centre. Theorem 4 says that the centre lies on the perpendicular bisector of every chord. If you have a broken piece of a round plate, draw two chords and their perpendicular bisectors. The two bisectors cross at the centre of the plate.

Using Pythagoras

Triangle $OMA$ has a right angle at $M$. With the radius $r$, the distance $d = OM$ and the chord $c = AB$, we get

Any one of the three numbers follows from the other two.

Example 1. A circle has radius $13\ \text{cm}$. A chord is $5\ \text{cm}$ from the centre. Find its length.

1. Half the chord. $AM^2 = 13^2 - 5^2 = 169 - 25 = 144$, so $AM = 12\ \text{cm}$.
2. The chord. $AB = 2 \times 12 = 24\ \text{cm}$.

Example 2. A chord of length $16\ \text{cm}$ lies in a circle of radius $10\ \text{cm}$. How far is it from the centre?

1. Half the chord. $AM = 8\ \text{cm}$.
2. The distance. $OM^2 = 10^2 - 8^2 = 36$, so $OM = 6\ \text{cm}$.

Some right triangles have all three sides whole numbers: $3, 4, 5$ and $5, 12, 13$ and $8, 15, 17$ and $7, 24, 25$. Doubling or tripling them gives more. These are quick to spot in chord problems.

Equal chords are equally far from the centre

Theorem 5. Equal chords of a circle are at equal distances from the centre.

Given. A circle with centre $O$, and chords $AB = CD$. $OM$ is perpendicular to $AB$ and $ON$ is perpendicular to $CD$.

To prove. $OM = ON$.

Proof. Take the triangles $OMA$ and $ONC$.

1. $AM = \frac{1}{2}AB$ and $CN = \frac{1}{2}CD$, by Theorem 3.
2. $AB = CD$, so $AM = CN$.
3. $OA = OC$, because they are radii. These are the hypotenuses.
4. $\angle OMA = \angle ONC = 90^\circ$.
5. So $\triangle OMA \cong \triangle ONC$, by the RHS rule.
6. Therefore $OM = ON$, by CPCT.

Theorem 6 (the converse). Chords of a circle that are at equal distances from the centre are equal.

Proof. Now $OM = ON$ is given. In the right triangles $OMA$ and $ONC$ we have $OA = OC$ and $OM = ON$. So the triangles are congruent by RHS, and $AM = CN$. Doubling, $AB = CD$.

Longer means nearer. Look at the simulation. When the distance $OM$ is $0$, the chord passes through the centre and is a diameter, the longest chord of all. As the chord moves away from the centre, it gets shorter.

Practise with new numbers

Try it in Python

This program finds the length of a chord at a given distance from the centre of a circle of radius $13$. It also checks the answer by finding $OA$ and $OB$ from coordinates.

Common mistakes

• Using the whole chord in Pythagoras. The right triangle has half the chord as a leg, not the whole chord.
• Using a slanting line as the distance. The distance is the perpendicular from the centre, not a radius to an end of the chord.
• Thinking that a longer chord is farther from the centre. It is the other way round: a longer chord is nearer.
• Forgetting to double. After finding half the chord, double it.

Remember these

• The perpendicular from the centre to a chord bisects the chord, and the line from the centre to the middle of a chord is perpendicular to it.
• $r^2 = d^2 + \left(\dfrac{c}{2}\right)^2$ links the radius, the distance and the chord.
• Equal chords are equally far from the centre, and chords equally far from the centre are equal.
• The nearer a chord is to the centre, the longer it is.

Check yourself

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

1. $OM$ is perpendicular to the chord $AB$, and $AM = 9\ \text{cm}$. Find $AB$.

2. A chord of a circle of radius $10\ \text{cm}$ is $6\ \text{cm}$ from the centre. How long is the chord?

3. Two chords of a circle are each $4\ \text{cm}$ from the centre. What can you say about their lengths?

4. Chord $AB$ is $2\ \text{cm}$ from the centre of a circle and chord $CD$ is $5\ \text{cm}$ from the centre. Which chord is longer?

5. Which chord of a circle is at distance $0$ from the centre?