What can we deduce if the loci traced out by two points touch?

What can we deduce if the loci traced out by two points touch?

Question

A point $P(x,y)$ moves so that $OP^2=\lambda y$, where $O$ is the origin and $\lambda$ is a constant. Show that the locus of $P$ is a circle. State the coordinates of the centre of the circle and the length of its radius.

Using the same axes, find the locus of a point $Q(x',y')$ which moves so that $QR^2=\lambda' y'$, where $R$ is the fixed point $(\alpha,0)$ and $\lambda'$ is a constant.

If the two loci touch each other, and $\alpha\ne 0$, prove that $\alpha^2=\lambda\lambda'$.

Suggestion

Could we sketch the point $P(x,y)$ and the line $OP$?

What’s the formula for the distance between two points?

Could we rearrange this into the standard form for a circle?

Could we draw a picture of this situation?

What’s the distance between the two centres here?

Interactive graph

Download GeoGebra file

The red circle here is $x^2 + y^2 = \lambda y$, while the green circle is $(x-\alpha)^2+y^2=\lambda'y$.

What happens as we vary $\lambda, \alpha$ and $\lambda'$? When do the circles touch?

Solution

We can apply Pythagoras’ Theorem to find the length

$OP^2$

. We have

$$OP^2=x^2+y^2.$$

Now we know that the length

$OP^2$

must be equal to

$\lambda y$

. We substitute this in and rearrange, giving

$$\begin{align*} OP^2=\lambda y =x^2+y^2 \\ x^2+y^2-\lambda y=0 \\ x^2+\left(y-\dfrac{\lambda}{2}\right)^2-\dfrac{\lambda^2}{4}=0 \\ x^2+\left(y-\dfrac{\lambda}{2}\right)^2=\dfrac{\lambda^2}{4}. \end{align*}$$

This tells us that points $P(x,y)$ such that $OP^2=\lambda y$ form a circle, and inspecting the equation above, we see that the circle has centre $\left(0,\dfrac{\lambda}{2}\right)$ and radius $\dfrac{\lambda}{2}$.

A point $Q(x',y')$

We see the length of the base of the triangle is $x'-\alpha$, if $x'>\alpha$, or $\alpha-x'$ if $x'<\alpha$.

As squaring both of these quantities gives the same result, we see that

$$\begin{align*} \lambda'y'=QR^2=(x'-\alpha)^2+y'^2 \\ (x'-\alpha)^2+\left(y'-\dfrac{\lambda'}{2}\right)^2=\dfrac{\lambda'^2}{4}, \end{align*}$$

that is, such points lie on the circle with centre $\left(\alpha,\dfrac{\lambda'}{2}\right)$, with radius $\dfrac{\lambda'}{2}$.

There is a right-angled triangle here with side lengths that we know.

Using Pythagoras’ theorem, we find

$$\left(\frac{\lambda}{2}+\frac{\lambda'}{2}\right)^2=\alpha^2+\left(\frac{\lambda}{2}-\frac{\lambda'}{2}\right)^2.$$

We could see this as finding (the distance between the centres of the circles$)^2$, using the formula for the distance between two points, which was originally derived from Pythagoras’s theorem.

Expanding the brackets gives

$$\frac{\lambda^2}{4}+\frac{\lambda'^2}{4}+\frac{\lambda\lambda'}{2}=\alpha^2+\frac{\lambda^2}{4}+\frac{\lambda'^2}{4}-\frac{\lambda\lambda'}{2}.$$

Rearranging, we have that

$$\alpha^2=\lambda\lambda'.$$