Let $A$ and $B$ be two points on the plane with $A B=7$. What is the set points $P$ such that $P A^{2}=P B^{2}-7$ ?
If we let $K$ be the point on $A B$ with $A K=4, B K=3$, then the answer is the line through $K$ perpendicular to $A B$. To see this, set $A=(0,0)$ and $B=(7,0)$. Then the points $P=(x, y)$ are exactly those satisfying

$$
(x-0)^{2}+(y-0)^{2}=(x-7)^{2}+(y-0)^{2}-7
$$

which rearranges to $9=-14 x+42$, id est $x=3$.