### 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 2

Following on from problem 1, we have two fixed points $A$ and $B$ in the plane with $AB=2a$, where $a>0$.

A point $P$ moves in the plane so that $\frac{PA}{PB}=\lambda,$ where $\lambda>0$.

We shall now assume that $\lambda\ne1$, so that the locus of $P$ is a circle.

What is the length of the tangent to this circle from the mid-point of $AB$, that is, the length $MN$ in the image below?

The radius and centre of the circle were found in the solution to problem 1.

This image shows the situation described in the case $\lambda=2$, where a coordinate system has been imposed and the line $CN$ has been drawn in.

Incidentally, there are actually two tangents from $M$ to the circle, but their lengths are equal. (Can you prove this?) We have only drawn one of them.

We note first that the tangent at $N$ is perpendicular to the radius at $N$, so the triangle $CMN$ is right-angled. We can therefore use Pythagoras’s Theorem: $CM^2=CN^2+MN^2.$

Since we know the coordinates of $C$ from the first part of the problem and $M$ is the midpoint of $AB$, we can easily work out the length $CM$. The length $CN$ is the known radius of the circle, and so we can deduce the length $CM$.

We use the nice coordinates approach from the first solution to perform our calculations: $A$ is at $(0,0)$ and $B$ is $(2a,0)$, as in the diagram above. Then $C$ is $\left(\dfrac{2\lambda^2 a}{\lambda^2-1}, 0\right)$ and the circle has radius $\dfrac{2\lambda a} {|\lambda^2-1|}$.

The midpoint $M$ has coordinates $(a,0)$, so \begin{align*} MN^2&=CM^2-CN^2\\ &=\left(\frac{2\lambda^2 a}{\lambda^2-1}-a\right)^2- \left(\frac{2\lambda a} {|\lambda^2-1|}\right)^2\\ &=\left(\frac{(\lambda^2+1)a}{\lambda^2-1}\right)^2- \left(\frac{2\lambda a}{\lambda^2-1}\right)^2\\ &=\frac{(\lambda^4+2\lambda^2+1)a^2-4\lambda^2a^2}{(\lambda^2-1)^2}\\ &=\frac{(\lambda^4-2\lambda^2+1)a^2}{(\lambda^2-1)^2}\\ &=a^2. \end{align*}

Therefore $MN=a$, which is $\frac{1}{2}AB$, independent of the choice of $\lambda$!

What is the locus of $N$ as $\lambda$ varies?

As $MN=a$ is a constant and $M$ is fixed, it follows that the locus of $N$ as $\lambda$ varies is (part of) the circle with diameter $AB$, which is very pretty.

Which parts of this circle are included in the locus of $N$ as $\lambda$ varies? Are there any parts which are not?