How to construct a circle in a the Poincare Disk model

Question

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?

Answer 1

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 $M$

Answer 2

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.