# Ford Circles

Can you find the link between these beautiful circle patterns and Farey Sequences?

## Problem

In the interactivity below, you can move the sliders to choose values for $a, b, c$ and $d$. The circles have centres $\left(\frac{a}{c},\frac1{2c^2}\right)$ and $\left(\frac{b}{d},\frac1{2d^2}\right)$, and radii $\frac1{2c^2}$ and $\frac1{2d^2}$.

When the two circles touch, they are coloured in blue.

Explore the interactivity and find some values of $a, b, c$ and $d$ that generate circles that touch each other.

In the problem Farey Neighbours, you are invited to explore the value of $ad-bc$ for two adjacent fractions $\frac bd$ and $\frac ac$ from any Farey Sequence.

Explore the value of $ad-bc$ for the touching circles that you have found.

What do you notice?

Can you prove that for any touching circles in the interactivity above, $|ad-bc|=1$?

Can you prove that, given two such circles which touch the $x$ axis at $\frac bd$ and $\frac ac$, the circle with centre $\left(\frac{a+b}{c+d},\frac1{2(c+d)^2}\right)$ and radius $\frac1{2(c+d)^2}$ is tangent to both circles?

## Getting Started

You might like to try the problem Baby Circle before attempting to prove the statements in this problem.

## Student Solutions

Navjot from Sherborne Qatar School sent us a lovely clear explanation:*explore the value of $ad-bc$ for the touching circles that you have found. What do you notice?*

I noticed that $ad-bc$ was always equal to $\pm 1$. This happened when the pair

of fractions were Farey neighbours.

For example, $\frac01, \frac 11: (0 \times 1) - (1 \times 1) = -1$

Or, $\frac 1{11}, \frac 1{12}: (1 \times 12) - (1 \times 11) =1$

In both of these cases, the two circles were tangent to each other.*Can you prove that for any touching circles in the interactivity above, $|adâˆ’bc|=1$?*

So it is given to us that the centre of circle $A$ is $\left(\frac ac , \frac1{2c^2}\right)$ with radius $\frac1{2c^2}$,

and the centre of circle B is $\left(\frac bd, \frac1{2d^2}\right)$ with radius

$\frac1{2d^2}$.

To show $|ad-bc| = 1$, I found the length between the centres of the two circles and equated it to the sum of the radii, which is also the length between the two points.

$|AB| =\sqrt{(\frac ac - \frac bd)^2 +(\frac1{2c^2} - \frac1{2d^2})^2)} = \frac1{2c^2} + \frac1{2d^2}$

$\Rightarrow \sqrt{\left(\frac{ad-bc}{cd}\right)^2 + \left(\frac{2d^2-2c^2}{4c^2d^2}\right)^2} = \frac{2d^2+2c^2}{4c^2d^2}$

$\Rightarrow \frac{(ad)^2 - 2abcd + (bc)^2}{(cd)^2}+ \frac{4d^4-8(dc)^2 + 4c^4}{16(cd)^4} = \frac{4d^4+8(dc)^2+4c^4}{16(cd)^4}$

$\Rightarrow \frac{(ad)^2 - 2abcd +(bc)^2}{(cd)^2}=\frac{(4d^4 + 8(dc)^2 + 4c^4) - (4d^4 - 8(dc)^2 + 4c^4)}{16(cd)^4}$

$\Rightarrow \frac{(ad-bc)^2}{(cd)^2} = \frac{16(cd)^2}{16(cd)^4}$

$\Rightarrow (ad-bc)^2=1$

$\Rightarrow |ad-bc|=1$.

We also received solutions from Sarith from Royal College in Sri Lanka, and Vignesh from Hymers College in the UK. You can read their solutions below:

Sarith's Solution

Vignesh's Solution

## Teachers' Resources

### Why do this problem?

This problem provides a beautiful extension to Farey Neighbours and encourages students to explore the connection between a geometrical pattern and a numerical sequence.

### Possible approach

It would be helpful if students were familiar with Farey Sequences and Farey Neighbours.

Start by demonstrating the first GeoGebra applet and defining Ford Circles:

"Ford Circles have centre $\left(\frac{p}{q},\frac1{2q^2}\right)$ and radius $\frac1{2q^2}$, where $\frac{p}{q}$ is a fraction in its simplest form (that is, $p$ and $q$ are coprime integers)."

Invite students to explore the second GeoGebra applet and challenge them to find some values of $a, b, c$ and $d$ that generate circles which touch. Record any that they find, and invite them to look for patterns, drawing attention to $ad-bc$ if it does not emerge from the class.

The two questions at the end of the problem are the key to the link between Ford Circles and Farey Sequences, and provide a good challenge in proof for older students:

- Can you prove that for any touching circles in the interactivity, $|ad-bc|=1$
- Can you prove that, given two such circles which touch the $x$ axis at $\frac bd$ and $\frac ac$, the circle with centre $\left(\frac{a+b}{c+d},\frac1{2(c+d)^2}\right)$ and radius $\frac1{2(c+d)^2}$ is tangent to both circles?

A diagram like this one might help:

$R=\frac{1}{2d^2}$ and $r=\frac{1}{2c^2}$.

The centre of the circle which touches the horizontal axis at $M$ is $(\frac{a+b}{c+d}, \frac{1}{2(c+d)^2})$.

### Key questions

What can you say about $R+r$ and $R-r$ if the circles centre $B$ and $A$ just touch each other?

Can you use Pythagoras theorem?

### Possible extension

Students may wish to read more about Ford Circles in this Wikipedia article.

### Possible support

As well as working on Farey Neighbours, Baby Circle would be a useful problem to try before attempting this challenging task.