# Locus of points bearing a fixed distance ratio between two other points in Riemannian manifold

Does the geometric locus of points bearing a fixed ratio between two other fixed points have a constant geodesic curvature?

Let $M$ be a 2-dimensional Riemannian manifold $x,y∈M.$ Is the set of points ${z |\,d(z,x)/d(z,y)}= \lambda$ comparable to the Apollonian circle in analogy for geodesic deviation?

Can it be generalized to higher-dimensional Riemannian manifolds?

I was wanting to be able to frame a similar question from here

• For $\mathbb R^3$ with the Euclidean metric the surface would be the sphere swept out by an Apollonian circle in a plane through $x$ and $y,$ rotated about the line through $x$ and $y.$ I think this generalizes to $\mathbb R^n.$ But you seem to be asking about a locus in a non-Euclidean metric space, which sounds like a different problem. The title doesn't seem to match the rest of the question. Mar 7, 2017 at 19:07
• Please suggest a better title, shall change it.Also how should the non-Euclidean metric space be. Mar 7, 2017 at 19:22
• Maybe just change "$\mathbb R^3$" to "Riemannian manifold" in the title. If I understand what you're trying to ask, the possible embedding of $M$ into $\mathbb R^3$ is irrelevant. But from the other question I guess you should also mention that $M$ has constant curvature. And that's about as far as my ability goes in attempting to address this question. I hope someone can answer it. Mar 7, 2017 at 20:28
• Thanks ! Shall try to grasp this slowly. Mar 7, 2017 at 22:24

Definition. Let $M$ be a Riemannian manifold, $\lambda>0$ is a number. Given two points $p,q\in M$ define the $\lambda$-bisector $B_\lambda(p,q)$ as the locus of points $x\in M$ such that $d(p,x)=\lambda d(q,x)$. (If $\lambda=1$, this is the ordinary bisector.)
Question. What is the shape of $B_\lambda(p,q)$? For instance, if $dim(M)=2$, is it true that for some $\lambda>0$, for all $p, q\in M$, $B_\lambda(p,q)$ is a constant curvature curve? (With curvature not prescribed in advance.)
If this is indeed the question, then a direct computation shows that for the flat metric on $R^2$ all $\lambda$-bisectors are circles, hence, they all have constant curvature. But if $M$ is the hyperbolic plane or the sphere (constant curvature $-1$ and $1$ respectively) then for generic points $p, q$ and all values $\lambda\ne 1$, $\lambda$-bisectors do not have constant curvature. (This is again a direct computation.) I very much doubt that there is a non-flat Riemannian metric on a surface such that for some $\lambda$ all $\lambda$-bisectors have constant curvature. Proving this, would require writing a serious research paper.