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?

    If we write out the equations for these two loci, we see: \[\frac{PA}{PB}=3\quad\text{and}\quad \frac{PA}{PB}=\frac{1}{3}.\] If we now take the reciprocal of the second equation, these become \[\frac{PA}{PB}=3\quad\text{and}\quad \frac{PB}{PA}=3.\] The equations are identical, except that \(A\) and \(B\) have swapped roles. So the loci are identical except for swapping \(A\) and \(B\), which can be achieved by reflecting in the perpendicular bisector of \(AB\).

    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:

    Triangle A, B, P with the lengths of its sides.
    Using these, we then see that \[\begin{equation} \label{eq:1} \frac{PA}{PB}=\frac{\sqrt{x^2 + y^2}}{\sqrt{(x-2a)^2 + y^2}} = \lambda. \end{equation}\]

    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.

    We rewrite \(\eqref{eq:1}\) with \(a=2\) and \(\lambda=3\), giving \[\begin{equation}\label{eq:frac1} \frac{\sqrt{x^2 + y^2}}{\sqrt{(x-4)^2 + y^2}} = 3. \end{equation}\] If we now square this to get rid of the square roots, we get \[\frac{x^2 + y^2}{(x-4)^2 + y^2} = 9.\] We can multiply through to lose the fractions, obtaining: \[x^2 + y^2 = 9 \bigl((x-4)^2 + y^2\bigr).\] It now makes sense to expand the inner brackets, giving \[x^2 + y^2 = 9 (x^2-8x+16 + y^2).\] We can then rearrange this to collect all of the terms on one side, giving \[8x^2+8y^2-72x+144 = 0,\] which can be divided by \(8\) to give \[\begin{equation}\label{eq:circ1} x^2+y^2-9x+18 = 0. \end{equation}\]

    This is the equation of a circle, so we’ve achieved what we set out to do after question 1!

    We also observe that every step of this argument can be reversed, so if the point \((x,y)\) lies on the circle, which means that it satisfies \(\eqref{eq:circ1}\), it will also satisfy \(\eqref{eq:frac1}\). (Note too that \((4,0)\) does not satisfy \(\eqref{eq:circ1}\), so there is no problem of dividing by zero in \(\eqref{eq:frac1}\) when working backwards.)

    To find the centre and radius of the circle, we complete the square to get \[\bigl(x-\tfrac{9}{2}\bigr)^2-\bigl(\tfrac{9}{2}\bigr)^2+y^2+18=0,\] so \[\bigl(x-\tfrac{9}{2}\bigr)^2+y^2=\tfrac{9}{4}.\]

    Therefore the locus is a circle with centre at \(\bigl(\tfrac{9}{2},0\bigr)\) and radius \(\tfrac{3}{2}\).

    To do the general case, we do exactly the same as in the specific example, we simply use \(a\) in place of \(2\) and \(\lambda\) in place of \(3\).

    So we again begin with \(\eqref{eq:1}\), square it and multiply through to get \[x^2 + y^2 = \lambda^2 \bigl((x-2a)^2 + y^2\bigr).\] We then expand and rearrange to find \[\begin{equation} \label{eq:2} (\lambda^2-1)(x^2+y^2)-4\lambda^2 ax + 4\lambda^2 a^2 = 0. \end{equation}\]

    If \(\lambda^2-1\ne0\), that is, \(\lambda\ne1\) (as \(\lambda>0\)), we can divide by \(\lambda^2-1\) to get \[x^2+y^2-\frac{4\lambda^2 ax}{\lambda^2-1} + \frac{4\lambda^2 a^2}{\lambda^2-1} = 0.\]

    This is the equation of a circle. As in the specific example above, the argument can be reversed, and so every point on the circle satisfies the original condition.

    To find the centre and radius of the circle, we now complete the square to get \[\begin{align*} \left(x-\frac{2\lambda^2 a}{\lambda^2-1}\right)^2+y^2 &=\frac{4\lambda^4 a^2}{(\lambda^2-1)^2}- \frac{4\lambda^2 a^2}{\lambda^2-1}\\ &=\frac{4\lambda^2 a^2}{(\lambda^2-1)^2}. \end{align*}\]

    Thus the locus is a circle with centre \(\left(\dfrac{2\lambda^2 a} {\lambda^2-1}, 0\right)\) and radius \(\dfrac{2\lambda a} {|\lambda^2-1|}\).

    Do check that this gives the same answer as we had above in the case \(a=2\) and \(\lambda=3\)!

    We need the absolute value in the formula for the radius to handle the case \(0<\lambda<1\): try substituting in \(a=2\) and \(\lambda=\frac{1}{3}\) to see why.

    If \(\lambda>1\), then \(B\) lies within the circle, whereas if \(\lambda<1\), then \(A\) lies within the circle, as shown in the applet above. (You might like to try showing this is the case from the algebra.)

    The only remaining case is when \(\lambda=1\), where we know that the answer is a straight line. We can obtain this from the algebra: when \(\lambda=1\), equation \(\eqref{eq:2}\) becomes \[-4ax + 4a^2 = 0.\] Dividing by \(4a\) (which is not zero) gives \(-x+a=0\) or equivalently \(x=a\): this is the perpendicular bisector of \(AB\).

    This makes sense, because the case \(\lambda=1\) gives \(PA=PB\), and the locus of all points equidistant from \(A\) and \(B\) is their perpendicular bisector.

    You might like to consider what happens as \(\lambda\) tends to zero or infinity, or when \(\lambda\) tends to \(1\). How does the algebra relate to what you have seen in the applet?

    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 \(\lambda\)?