### Geometry of Equations

Many ways problem

# The circle of Apollonius... coordinate edition Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

## Solution 1

Two fixed points $A$ and $B$ lie in the plane, and the distance between them is $AB=2a$, where $a>0$.

A point $P$ moves in the plane so that the ratio of its distances from $A$ and $B$ is constant: $\frac{PA}{PB}=\lambda,$ where $\lambda>0$.

1. Can you sketch the locus of the point $P$ for different values of $\lambda$?

Here are some values of $\lambda$ you might consider:

• $\lambda=1$

• $\lambda=3$

• $\lambda=\frac{1}{3}$

This GeoGebra applet shows the locus of $P$. The points $A$ and $B$ are movable, and the value of $\lambda$ can be adjusted using the slider.

You can use this applet to check your sketches for the cases $\lambda=1$, $\lambda=3$ and $\lambda=\frac{1}{3}$.

You may have noticed that the loci for the cases $\lambda=3$ and $\lambda=\frac{1}{3}$ look essentially the same as each other, just reflected. Why should this be the case?

We’d like to show that the locus really is a circle, which brings us to question 2.

2. Using Cartesian coordinates, work out (the equation of) the locus of $P$.

You may find it more straightforward to first work with specific values of $a$ and $\lambda$, say $a=2$ and $\lambda=3$.

Suggestion: The question has not specified the coordinates of the points $A$ or $B$, so what options do we have?

In particular, what would be “nice” coordinates to choose for the points $A$ and $B$?

As we are going to be using coordinates in this problem, and are looking for the locus of $P$, we will let $P$ have coordinates $(x,y)$.

We could let $A$ be at $(q,r)$ and $B$ be at $(s,t)$, for example. However, this is likely to make our algebra very messy, having so many letters around.

A better approach is to choose “nice” coordinates for $A$ and $B$. Let’s take $A$ to be at the origin, and $B$ to be on the $x$-axis. This way, $A=(0,0)$ and $B=(2a,0)$, and most of the coordinates involved are zero.

Another good choice would be $A=(-a,0)$ and $B=(a,0)$, so that the points are symmetrically placed about the origin. We will go with the first option in this solution, but you might like to have a go with this alternative choice and compare them.

We let $P$ have the coordinates $(x,y)$, and then apply Pythagoras to find $PA=\sqrt{x^2 + y^2}\ \text{and}\ PB=\sqrt{(x-2a)^2 + y^2},$ as seen in this sketch:

Using these, we then see that $$$\label{eq:1} \frac{PA}{PB}=\frac{\sqrt{x^2 + y^2}}{\sqrt{(x-2a)^2 + y^2}} = \lambda.$$$

We will give the rest of the solution first for the specific example suggested, $a=2$ and $\lambda=3$, and then for the general case.

If you did try this question with coordinates such as $(q,r)$ for $A$ and $(s,t)$ for $B$, you will have found that the algebra was very messy indeed.

It is always worth asking oneself what options one has when choosing coordinates: a good choice can have huge advantages.

In this situation, we were free to choose both the origin and the orientation of the axes, so we were able to choose useful ones.

The locus of $P$ is known as the circle of Apollonius. See the Historical background for this problem to read more about Apollonius and this circle.

A further question to consider: Can we go backwards?

In other words, if we begin with a specified circle, is it always possible to find two points $A$ and $B$ and a value $\lambda>0$ so that the given circle is the locus of the point $P$ satisfying $PA/PB=\lambda$?

If it is possible, is there any choice in how we can choose $A$, $B$ and $\lambda$?