find the locus of points M

Question

find the locus of points M, the difference between the squares of the distances from which to two given points A and B is equal to a given value C, At which C does the problem have a solution?

Anyone know how to solve this?

Answer 1

This locus is a line perpendicular to $AB$. For proving it we'll prove the following:

In Figure1, $AB$ is perpendicular to $CD$ iff $AC^2-AD^2=BC^2-BD^2$

Proof: Assume that $H_a,H_b$ are the feet of altitudes from $A,B$ to $CD$, respectively. By pythagorean theorem,
$$AC^2-AD^2=AH_a^2+CH_a^2-AH_a^2-DH_a^2=CH_a^2-DH_a^2$$
Similarly, $BC^2-BD^2=CH_b^2-DH_b^2$. Therefore it suffices to prove that $H_a\equiv H_b$ iff $CH_a^2-DH_a^2=CH_b^2-DH_b^2$ and this is trivial since:
$$CH_a^2-DH_a^2=CH_b^2-DH_b^2 \implies (CH_a+DH_a)(CH_a-DH_a)=(CH_b+DH_b)(CH_b-DH_b)$$
But since $CH_a+DH_a=CH_b+DH_b$,
$$CH_a-DH_a=CH_b-DH_b$$
But since sum of them are also equal, $CH_a=CH_b$ and hence the result.