If $A=(-4,0)$ and $B=(4,0)$, what is the locus of points $P$ such that $|AP-BP|=16$? Does it even exist?

Question

I am stuck in this question for about a week:

If there are points $A$ and $B$ such that $A(-4,0)$ and $B(4,0)$ then what is the locus of points $P$ such that $|AP-BP|=16$?

I think this is a trick question. But I do not think this is possible. If $APB$ is a triangle, then $|AP-BP|\leq 8$. So $APB$ cannot form a triangle.

But if my answer is incorrect, and such points $P$ exist, what will the points $P$ be?

Answer 1

You are correct - the reverse triangle inequality implies that $$|d(A,P)-d(B,P)|\leq d(A,B)=8$$ so there will be no points $P\in\mathbb{R}^2$ such that $|d(A,P)-d(B,P)|=16$. Here is a nice contour plot of the values of $|d(A,P)-d(B,P)|$ from Mathematica that illustrates this:

A = {-4, 0}; B = {4, 0};

ContourPlot[ Abs[EuclideanDistance[A, {x, y}] - EuclideanDistance[B, {x, y}]],
 {x, -15, 15}, {y, -15, 15}, ContourLabels ->
 (Text[Style[Framed[#3, RoundingRadius -> 10, Background -> White],
 Directive[12], FontFamily -> "Helvetica"], {#1, #2}] &), 
 LabelStyle -> Directive[12, Bold, FontFamily -> "Helvetica"]]

Answer 2

AP-BP is the vector from A to B, which has length $8$. The locus is the null set.