# How to draw Hyperbolic Circle in the Klein-Beltrami Disk

I'd like to know how to draw an hyperbolic circle in the Klein-Beltrami model just knowing its center and its radius, or a center and a point on it.

Apparently, this hyperbolic circles are represented by ellipses as shown on this Wikipedia picture.

I guess the (euclidean) line made by the focal points are perpendicular to the (euclidean) line made by the centers of the disk and of the circle. Also, the center of the hyperbolic circle should not be aligned with the focal points of the euclidean ellipse since it should be farthest.

Any starting point might help and a formula would be very welcomed.

• Think about a circle whose center is near the edge of the disc but that contains the center point inside it. It can't be as you think it is. The "center" of the circle/ellipse must change. – Xoff Oct 18 '17 at 10:24

So, a segment and a circle center is given in the Klein model:

and the task is to construct points, say $D$, for which $CD$ is congruent to $AB$ -- congruent in the hyperbolic sense.

This is how one can construct as many such points, as she wants.

Step 1.

Draw the straight line containing $AB$ and draw a movable straight line through $C$ as shown below:

Here $H$, the "handle" is the point that, if moved, will turn the right straight around the circle center $C$. Note that the part of the white lines within the circle are the hyperbolic straights. Also, in the figure above four yellow lines are drawn. I just hope that I don't have to further explain the construction of the yellow lines.

Step 2.

The next step is to connect the intersection points of the yellow lines with a straight line (red). Then construct the blue line as shown below. Finally, construct the black line that will give point $D$, a point of the circle we try to construct.

Step 3.

Now, move the handle around $C$. Then $D$ will trace the hyperbolic circle.

Note that such a dynamic construction and tracing can be done by a dynamic geometry software like Geogebra or Cinderella...

But, if tracing is not allowed then think of the following fact: five points determine an ellipse. So, it would be enough to construct five $D$'s.

• I didn't get everything you said. What is hyperbolic congruence for segments ? Can I choose H, the handle, to be any arbitrary point ? Knowing that an hyperbolic circle which center coincides with the center of the disk is represented by an euclidean circle, it's far easier for me to apply a Möbius transform to "pin" the circle, choose 5 points on it and then applying the reverse Möbius transform. Maybe I should say that in the question ? – Kii Oct 18 '17 at 12:14
• The hyperbolic congruence is given by the construction: yellow lines, red line, blue line then the black line... This is because this construction preserves the cross ratio. $H$, the handle can be anywhere (but at $C$) If you try this construction with $C$ in the middle of the Klein circle then you will get a circle. I don't know what to do with the Möbius thing in the Klein model. I would suggest that you try to understand why the construction preserves the cross ratio. THAT is the real understanding of hyperbolic geometry within the Klein circle. I would not experiment with formulae. – zoli Oct 18 '17 at 12:34
• I answered my question with the solution I used. It worked like a charm. Thanks for your answer anyway. I implemented the drawing of circles in my hyperbolic geometry software now. Tell me if you're interested in. – Kii Dec 1 '17 at 14:39
• @Kii Yes, I am interested. (I have an hyprbolic software too.) – zoli Dec 1 '17 at 20:21
• You can check it out there : framagit.org/Kii/hyperbolic-browser Feedbacks are welcomed ! – Kii Dec 3 '17 at 10:11

Ok, so I contacted the author of the Wikipedia picture and he gave me the method and formulas he used to draw the hyperbolic circles as euclidean ellipses.

To draw them, he calculated the ellipse thanks to its major-axis and minor-axis. Here's a translation of the formulas.

Point A (x, y) being an element of the Beltrami-Klein Disk, the hyperbolic circle with center $A$ and radius of hyperbolic length $r$ is define as follow:

The euclidean center C is located at:

$C = \frac{A}{cosh(r)^2 (1 - A^2) + A^2 }$

The length a of the semi-axis in the OA direction is :

$a = \frac{cosh(r)sinh(r)(1 - A^2)}{cosh(r)^2 (1 - A^2) + A^2 }$

The length b of the semi-axis orthogonal to OA and going through C is :

$b = \frac{sinh(r) \sqrt{1 - A^2}}{\sqrt{cosh(r)^2 (1 - A^2) + A^2}}$

In the preceding formulas, $cosh(r)$ is the hyperbolic cosine and $sinh(r)$ is the hyperbolic sine. Also, we define $A^2 = x^2 + y^2$