Exercise 1318Geometry › Analytic geometryOlympiadAdded 5 Nov 2026

The Apollonius circle

#locus#Apollonius circle#coordinates#angle bisector

Statement

Let $A$ and $B$ be two points at distance $d>0$ , and let $k>0$ , $k\ne1$ .

Prove that the locus of the points $P$ with $PA=k\cdot PB$ is a circle. Find its centre and radius.
Let $E$ and $F$ be the two points of line $AB$ on this locus. Prove that the centre of the circle is the midpoint of $EF$ .
Let $P$ be a point of the locus not on line $AB$ . Prove that $PE$ and $PF$ are the internal and external bisectors of $\angle APB$ .

Hint

Put $A=(0,0)$ and $B=(d,0)$ and square the condition $PA=kPB$ .

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Solution

1. For $P=(x,y)$ : $x^2+y^2=k^2\left((x-d)^2+y^2\right)$ , i.e.
$(k^2-1)(x^2+y^2)-2k^2d\,x+k^2d^2=0\ \Longleftrightarrow\ \left(x-\frac{k^2d}{k^2-1}\right)^2+y^2=\frac{k^2d^2}{(k^2-1)^2}>0.$
So the locus is the circle with centre $\left(\dfrac{k^2d}{k^2-1},0\right)$ and radius $\dfrac{kd}{|k^2-1|}$ .

2. On line $AB$ (the $x$ -axis) the circle meets the axis at $x=\frac{k^2d\pm kd}{k^2-1}$ , that is at $x=\frac{kd}{k-1}$ and $x=\frac{kd}{k+1}$ . These are the points $F$ and $E$ with $\frac{EA}{EB}=\frac{FA}{FB}=k$ . Their midpoint is $\frac{k^2d}{k^2-1}$ , the centre. Hence $EF$ is a diameter.

3. Since $PA/PB=k=EA/EB$ , the point $E$ lies on segment $AB$ and divides it in the ratio of $PA$ to $PB$ , so by the converse of the angle bisector theorem $PE$ bisects $\angle APB$ internally. Similarly $F$ lies outside the segment on line $AB$ with $FA/FB=PA/PB$ , so $PF$ is the external bisector.

Remark. For $k=1$ the locus degenerates to the perpendicular bisector.

Lesson. A locus defined by a distance relation is found by coordinates and completing the square; then identify the special points on the axis.

The Apollonius circle

Statement

Solution

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