Review question

# One distance is double another; can we find a locus? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8080

## Solution

Given that $A$ is the point $(0,3)$ and $B$ is the point $(0,-3)$, a point $P(x,y)$ moves so that $PA=2PB$. Show that the equation of the locus of $P$ is $x^2 + y^2 +10y + 9 = 0.$

The point $P$ has the coordinates $(x,y)$, so we can apply Pythagoras’ theorem to find the distances \begin{align*} PA&=\sqrt{x^2 + (y-3)^2}\\ PB&=\sqrt{x^2 + (y+3)^2}, \end{align*}

as seen in this sketch:

We can now insert the expressions for $PA$ and $PB$ into the constraint $PA=2PB$, giving $\sqrt{x^2 + (y-3)^2} = 2\sqrt{x^2 + (y+3)^2}.$

Next, we square both sides and expand the brackets. We have

\begin{align*} &&\sqrt{x^2 + (y-3)^2} &= 2\sqrt{x^2 + (y+3)^2}&&\quad\\ \implies\quad&& x^2 + (y-3)^2 &= 4x^2 + 4(y+3)^2\\ \implies\quad&& x^2 + y^2-6y+9&=4x^2+4y^2+24y+36. \end{align*}

Collecting like terms gives $3x^2 + 3y^2 + 30y + 27 = 0$ which, on dividing by $3$, gives the equation of the locus of $P$ in the form requested, namely $x^2 + y^2 +10y + 9 = 0$.

A more general problem with detailed suggestions and solutions can be found in The Circle of Apollonius.