Exercise 641Geometry › Analytic geometryM1Added 5 Nov 2026

Two classical locus problems

#locus#coordinates#perpendicular#circle with a given diameter

Statement

A circle has centre $O$ and radius $R$ , and $P\ne O$ is a point inside the circle. What is the locus of the midpoints of all the chords passing through $P$ ?
Let $A\ne B$ be two points and $k$ a real number. What is the locus of the points $X$ such that $XA^2-XB^2=k$ ?

Hint

For part 1, the line from the centre to the midpoint of a chord is perpendicular to the chord. For part 2, use coordinates with $A$ and $B$ on the $x$ -axis.

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Solution

1. Let $M$ be the midpoint of a chord through $P$ . If $M\ne O$ and $M\ne P$ , then $OM\perp$ chord, and the chord passes through $P$ , so $OM\perp MP$ : $M$ is on the circle with diameter $OP$ . The endpoints $M=P$ and $M=O$ are also on this circle. Conversely, every point of the circle with diameter $OP$ arises this way.

Answer. The locus is the whole circle with diameter $OP$ .

2. Put $A=(0,0)$ and $B=(d,0)$ with $d=AB>0$ . For $X=(x,y)$ ,
$XA^2-XB^2=2dx-d^2.$
This equals $k$ iff $x=\dfrac{k+d^2}{2d}$ .

Answer. The locus is a line perpendicular to $AB$ at abscissa $\frac{k+d^2}{2d}$ (for $k=0$ , the perpendicular bisector of $AB$ ).

Remark. Part 2 is the key lemma behind radical axes.

Lesson. A right angle gives a circle (Thales); a difference of squared distances gives a line.

Two classical locus problems

Statement

Solution

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