What is the solution set for point $X$ if $|AX|^2+ |BX|^2=c$?

Question

Let there be distinct points $A$ and $B$ and a given $c\gt0$. What is the solution set for point $X$ if $|AX|^2+ |BX|^2=c$?

I started solving this problem as: $\overrightarrow{AB}=\overrightarrow{AX}-\overrightarrow{BX}\implies(\overrightarrow{AB})^2=(\overrightarrow{AX}-\overrightarrow{BX})^2\implies|AB|^2=|AX|^2-2\overrightarrow{AX}\cdot\overrightarrow{BX}+|BX|^2 $
$\implies|AB|^2+2\overrightarrow{AX}\cdot\overrightarrow{BX}=|AX|^2+|BX|^2$

It is clear that $|AB|^2+2\overrightarrow{AX}\cdot\overrightarrow{BX}=c$

But from here on I am completely lost how to determine the solution set for point X.

SOS. I need serious help. Any tips?

Answer 1

Let $M$ be the midpoint of the line segment $AB$. Show (e.g., using Stewart's Theorem) that $$AX^2+BX^2=2MX^2+\frac{AB^2}{2}\,.$$ This result is known as Apollonius's Theorem.

Thus, if $c>\dfrac{AB^2}{2}$, then the locus of the point $X$ is a circle centered at $M$ with radius $\sqrt{\dfrac{c}{2}-\dfrac{AB^2}{4}}$. If $c=\dfrac{AB^2}{2}$, then $X=M$. Otherwise, $X$ does not exist.

Answer 2

Let $z$, $a$ and $b$ to represent the points $X$, $A$ and $B$, respectively, in the complex plane. From the given $|AX|^2+ |BX|^2=c$, we have

$$(z-a)(\bar z - \bar a) + (z-b)(\bar z - \bar b )= c$$

or,

$$|z|^2 - \frac{a+b}2\bar z - \frac{\bar a+\bar b}2z = \frac{c-|a|^2-|b|^2}2$$

which is equivalent to

$$|z - \frac{a+b}2|^2 = \frac{2c-|a-b|^2}4 = r^2$$

Thus, $|AX|^2+ |BX|^2=c$ represents a set on a circle with center $\frac{a+b}2$ and radius $r$ if $2c \ge |a-b|^2$; empty otherwise.