Review question

# What can we deduce if the loci traced out by two points touch? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8509

## Solution

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.

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 $\dfrac{\lambda}{2}$.

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.

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 $\dfrac{\lambda'}{2}$.

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

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

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