consider the Poincare upper half-plane model of hyperbolic plane $\mathbb{H}^2$ and a hyperbolic line $\ell\subset \mathbb{H}^2$ (or geodesic if you want). I would like to visualize the set of points equidistant to the given geodesic. For instance, consider the geodesic $\ell:=\lbrace (x,y)\in\mathbb{H}^2 \vert x^2+y^2=r^2 \rbrace$, where $r>0$ is fixed. How does the set of points $A\in\mathbb{H}^2$ with fixed distance $d>0$ to the line $\ell$ look like?

Of course some qualitatively properties are obvious. From the euclidean eye it will not look equidistant to $\ell$, but it will get very close to $\ell$ at "the endpoints" $(r,0), (-r,0)$ and will have something like a hill around $(0,r)$.

Is there a more effective way/approach to get a better picture of this set?

Further, consider another line $\tilde{\ell}:=\lbrace (x,y)\in\mathbb{H}^2 \vert x^2+y^2=R^2 \rbrace$, with $R>r.$ I would like to determine for which distances $d$ the two sets of points which have distance equal to $d$ to $\ell,\tilde{\ell},$ respectively, will meet or will be disjoint?

I would be very glad, if someone could help me.


The equidistant lines to $\ell$ are exactly the Euclidean circles passing through the ideal endpoints $\xi_- = (-r,0)$, $\xi_+ = (+r,0)$ of the line $\ell$ (intersecting those circles with the upper half-plane, of course). The line $\ell$ is, of course, the unique such circle that meets the real line at angle $\pi/2$. The other equidistant lines meet the real line at angles ranging between $0$ and $\pi/2$.

If you wish to construct the equidistant line to $\ell$ with an explicit distance $d$, once you find a single point $p$ with distance $d$, then the equidistant line is just the Euclidean circle passing through the three points $\xi_-,p,\xi_+$. For example, the point $q=(0,r)$ lies on $\ell$, the vertical line $x=0$ is perpendicular to $\ell$, and so we can take $p=(0,re^d)$.

To answer your "further" question, again by consideration of the vertical line through $(0,r)$ one sees that the distance between $\ell$ and $\tilde \ell$ equals $\log(R/r)$. So the equidistant sets are disjoint if $d<\frac{1}{2} \log(R/r)$, their intersection is a point if $d=\frac{1}{2} \log(R/r)$, and their intersection is two points if $d>\frac{1}{2} \log(R/r)$.

Added: To prove that equidistant lines have the form described, it is easier first to consider the equidistant lines to the hyperbolic line $x=0$. The line $x=0$ is invariant under the group of isometries $f_t(z)=tz$, $t > 0$. So for each point $p=(x,y)$, the distance from $p$ to the line $x=0$ equals the distance from $f_t(p)=(tx,ty)$ to the line $x=0$. In other words, the lines equidistant from $x=0$ are the lines through $(0,0)$, i.e. the lines of the form $x=my$.

Now to get the form of equidistant lines to the hyperbolic line $x^2 + y^2 = r^2$, do the case $r=1$ first, then scale. For the case $r=1$, first find the coefficients $a,b,c,d \in \mathbb{R}$ of a fractional linear transformation $T(z) = \frac{az+b}{cz+d}$ which takes the line $x=0$ to the line $x^2+y^2=1$, i.e. which takes the ideal points $(0,0)$ and $\infty$ to the ideal points $(-1,0)$ and $(+1,0)$. Then compute the image under $T$ of the line $x=my$. You will obtain a circle passing through $(-1,0)$ and $(+1,0)$.

  • $\begingroup$ Thank you very much for your complete answer! How do you know the equidistant points will form those circle arcs which you described? $\endgroup$ – asd Aug 6 '15 at 9:08
  • 1
    $\begingroup$ @asd: I added a justification. $\endgroup$ – Lee Mosher Aug 6 '15 at 13:30

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