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How can I construct an circle with centre C going trough point P in a Poincare disk?.

I found an script of how to do it in the "Poincaré Disk Model of Hyperbolic Geometry"toolkit from the geometers sketchpad,

http://www.dynamicgeometry.com/General_Resources/Advanced_Sketch_Gallery.html and http://www.dynamicgeometry.com/documents/advancedSketchGallery/Poincare_Disk.gsp

But the construction is long (47 steps) and gardled (or at least I cannot understand it)

are there easier or, more importandly, easier to understand ways to do this construction?

Willemien
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2 Answers2

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Do you know how to construct a circle inversion? If so, draw any two hyperbolic lines $g_1$ and $g_2$ through $C$ but not through $P$. Invert $P$ in $g_1$ to obtain $P_1$ and also in $g_2$ to obtain $P_2$. The Euclidean circle through $P,P_1,P_2$ is the hyperbolic circle. This is because a point on the circle, reflected on a diameter of the circle, will again lie on the circle.

If you don't know how to construct a hyperbolic line through $C$, simply invert $C$ in the unit circle (i.e. the boundary of the model) to obtain $C'$, then any circle through $C$ and $C'$ will be orthogonal to the unit circle, and hence a hyperbolic line.

Since all of the above is based on inversion in a circle, here is the construction which I'd use for this. It is an application of the standard harmonic set construction from projective geometry, based on the fact that the cross ratio $\operatorname{cr}(1,-1;x,\frac1x)=-1$. $P$ is reflected in the circle to obtain $P'$. $C$ is chosen arbitrarily, $D$ arbitrarily on $PC$. The rest follows. As alternatives, you may consider this image on Wikipedia. On the other hand, this question about a ruler-only construction is essentially the same construction I am using, even though it looks different due to the different choice for arbitrary points. The explanations as to why this works might be of interest, though.

Figure

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  • Thanks I know how to creaate (random) hyperbolic lines trough a point but not how to " Invert (reflect?) $P$ in $g_1$ to obtain $P_1$ " can you elaborate on this? – Willemien Jul 12 '14 at 21:09
  • @Willemien: See this image from Wikipedia as well as this question about a ruler-only construction for details. Or wait a second, I'll update my post. – MvG Jul 12 '14 at 21:24
  • I think I found a shorter method can you check if it is correct? - 1 Point D is the inverse point of C to the P-disk. - 2 point E is the midpoint of C and D. - 3 circle $C_1$ has centre E and goes trough C. - 4 point F is the midpoint of E and P. - 5 circle $C_2$ has centre F and goes trough P. - 6 Point Q is the midpoint of the two points where circle $C_1$ and circle $C_2$ meet. - 7 point R is the midpoint of Q and P. - 8 line l is perpendicular PQ trough R. - 9 point M is where line l and Pdisk-centre - A meet. - 10 the circle we need has centre M and goes to P. - is this correct? – Willemien Jul 12 '14 at 21:58
  • @Willemien: Sorry for not to get your aim of the question. I deleted it and a "+1" for this nice answer. :) – Mikasa Jul 13 '14 at 08:23
  • @Willemien: Yes, I can confirm that your construction works. Feel free to poste that as an answer to your own question, perhaps with some ideas as to why this works. – MvG Jul 13 '14 at 08:48
  • :) no it doesn't, After my comment I realised it doesn't work when $ P$ is on or inside circle $C_1$, so i am back to the sketchpad – Willemien Jul 14 '14 at 07:21
  • @Willemien: Well, $Q$ is what you get if you invert $P$ in $C_1$, so if you know how to do circle inversion, you can use that even if $P$ is inside $C_1$. If $P$ is on $C_1$ then $P$ and $Q$ will coincide, in which case their midpoint $R$ will be equal to both. By the way, I forgot to mention that your step 9 is unclear, particularly since you mention a point $A$ which I assume should be $C$. – MvG Jul 14 '14 at 07:40
  • @Willemien: Realizing that $P$ and $Q$ relate by an inversion, I now also understand why your approach works. You take a pretty arbitrary line though $C$, namely $C_1$, then reflect $P$ in that. These steps are similar to my solution. But then you don't use a second arbitrary line, but instead intersect with the diameter through $C$. One more note: to handle the case of $P=Q=R$ you should define $l$ not as prependicular to $PQ$ but rather perpendicular to $PE$. – MvG Jul 14 '14 at 07:45
  • Would you say that an inversion over (the circle of) an hyperbolic line equals an hyperbolic reflection over that line? (I needed to be able to make a circle to do reflexions over lines, but now I am wondering is the inversion not allready an reflexion) – Willemien Jul 14 '14 at 09:34
  • @Willemien: Yes, Euclidean circle inversion is hyperbolic reflection in a line. At least if the circle is orthogonal to the boundary of the disk. I'd consider it to be a more fundamental concept than that of a hyperbolic circle, in particular of a hyperbolic circle with given center and point on the boundary. Try to avoid dealing with hyperbiolic circle centers unless you have to. – MvG Jul 14 '14 at 09:49
  • I didn't know that inversion is a reflection, (really, I checked my textbooks , none did mention it) therefore I was thinking i needed a circle to do reflections, still knowing how to make an hyperbolic circle is good knowledge – Willemien Jul 14 '14 at 11:33
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With the help of https://math.stackexchange.com/users/35416/mvg Thanks !!) I found a shorter method:

  1. Point $D$ is the inverse point of C where the P-disk is the reference circle.
  2. point $E$ is the midpoint of C and D.
  3. circle $C_1$ has centre E and goes trough C.
  4. Point $Q$ is the inverse point of P where circle C1 is the reference circle.
  5. point $R$ is the midpoint of Q and P.
  6. line $l$ is perpendicular PE trough R.
  7. point $M$ is where line $l$ and the line Pdisk-centre - A meet
  8. the circle we need has centre M and goes to P.

Done

Explanation:

Circle $C_1$ is an hyperbolic line trough $C$ (it is ortogonal to the boundary circle)

Point Q is the hyperbolic reflection of point P over hyperbolic line $C_1$

line $l$ is the euclidean equidistant line of P and Q

Line P-disk-centre - A is also an a hyperbolic line trough $C$ and a line trough the circle centre. so the euclidean circle centre point is at point $M$

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