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Suppose I have a point $P = (x_1,y_1)$ in the Poincaré disk model. How do I rotate it about another point $Q = (x_2,y_2) \neq(0,0)$ by a Euclidean angle $\alpha$?

If $Q = (0,0)$ this is simple, just apply a normal Euclidean rotation, but I can't figure out how to do it for another point.

Do I need to apply a translation of the disk to move $Q$ to $(0,0)$, apply the rotation, then translate back? Or is there a direct way?

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  • $\begingroup$ I think your idea "Do I need to apply a translation of the disk to move Q to (0,0), apply the rotation, then translate back?" is easiest except use a reflection instead of a translation. (but that is a minor point I guess) $\endgroup$
    – Willemien
    Feb 10, 2016 at 13:27
  • $\begingroup$ The method of your last paragraph is certainly the correct one. If you think of it as too "indirect", simply write down the formulas for the maps involved in that method, and compose them to get a formula for the rotation. $\endgroup$
    – Lee Mosher
    Feb 12, 2016 at 14:19

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Found a better way :)

The idea is:

  • Construct the hyperbolic ray $Qr$ that is ray $QP$ $\angle \alpha$ rotated around $Q$

  • Construct circle $e$ (hyperbolicly) centered at $Q$ through $P$

  • The point you are looking for is the intersection of $e$ and $Qr$

The construction is a bit complicated:

In the construction below the elements are Euclidean elements (so lines and rays are straight Euclidean lines) also extent the elements to outside the boundary circle.

  • let $O$ be the centre of the Poincare disk

  • Draw circle $c$ that when inside the Poincare disk represents the hyperbolic line trough P and Q

  • Point $C$ be the centre of $c$

  • Point $C^1$ is point $C$ rotated $\angle \alpha$ around point $Q$

  • Draw ray $q$ from $O$ through $Q$

  • Draw line $d$ through $C$ perpendicular to $q$ (sometimes you allready have constructed this line to find point $C$)

  • Draw line $s$ from $Q$ through $C^1$

  • Point $R$ is the intersection of $d$ and $s$ (so that $\angle CQR = \angle \alpha$ and $C$ and $R$ are on $d$ )

  • Draw circle $r$ centered around $R$ through $Q$ (part of the arc inside the Poincare disk is the ray we wanted to construct)

Then circle $e$ (hyperbolicly) centered at $Q$ through $P$

  • Draw segment $CP$

  • Draw line $p$ through $P$ perpendicular to $CP$

  • Point $E$ is the intersection of $p$ and $q$

  • Draw circle $e$ centered around $E$ through $P$ (this is the circle hyperbolicly centered around $Q$ through $P$ )

  • One of the intersections of $e$ and $r$ is the point you are looking for.

DONE :)

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  • $\begingroup$ Thank you, with a bit of work that did the trick :) $\endgroup$
    – Lewy Blue
    Feb 13, 2016 at 17:58
  • $\begingroup$ What is a hyperbolic Ray ?? - Never seen the word Ray used in this context. $\endgroup$ Dec 5, 2023 at 20:59

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