Firstly, please note that the related question can also be found at mathoverflow.

The question is stated as following:

In Euclidean Geometry, we know that from a given point there is an unique line perpendicular to a given line. In constant negative curvature spaces (hyperbolic geometry), from the answers over mathoverflow, I know that there also exists an unique (geodesic) line which pass a given point and perpendicular to a given line.

After a second thought of my question, I also found a method to prove it. Since we know that there are several models for hyperbolic geometry, in the projective disk model, the proof should be almost apparent. But in the conformal disk model, the proof is not so easy to reach.

Now since the answer is positive, I asked to exactly work it out in conformal disk model. More exactly,

diagram goes here As the above figure shows, disk A is the conformal model disk. J is the given point, and CD is the given line; I asked how to get the line GF, such that J is on it and GF perpendicular to CD, ie., $\angle M=\pi/2$.

P.S.The step should only be completed by compass and straightedge.


2 Answers 2


This is about compass and straightedge. Draw the half-line beginning at A and passing through J, meeting the circle at a point we will call, say, R. Let the distance between A and R be called $r.$ Let the distance between A and J be called $j.$ We need to draw the point W along the ray AJ with distance from A exactly $$ \beta = \frac{r^2 + j^2}{2j}.$$ There is a standard construction for distance $\frac{r^2}{j},$ called the polar. Draw the line perpendicular to AJ through J. This line intersects the circle twice, pick one, call it T, draw the tangent to the circle at that point T (perpendicular to the radius through T), find the point X where this tangent intersects AJ. This is the point at distance $\frac{r^2}{j}$ from A. Along AJ, draw the point farther out than X by exactly $j,$ call this Y. Y is at distance $ \frac{r^2 + j^2}{j}$ from A. Finally, draw the midpoint Z of the segment AY. The distance between A and Z is $ \beta = \frac{r^2 + j^2}{2j}.$ Let line $\Lambda$ be the perpendicular to AZ through Z. Note that $\Lambda$ is not the polar of J with respect to the circle. Instead, every circle with a center on $\Lambda$ that passes through J will be orthogonal to the circle passing through your C, G, D, F. I do not know a name for this line.

Let line $\Delta$ be the line through C and D. This is called the "chordal." Also called the "power line" or "radical axis." It is, in Dorrie's prose, the "locus of the centers of all circles that are perpendicular to two given circles." In our case, the two given circles intersect, in the points C,D, and we have "The chordal of two circles that intersect is the secant of intersection."

Lines $\Lambda$ and $\Delta$ intersect in a point, call it H. Draw the circular arc with center at H and passing through J. Done.

This material is very old and hard to find. One of the best books giving at least part of this is 100 Great Problems of Elementary Mathematics by Heinrich Dorrie. The trouble these days is that architectural and engineering drafting is done by CAD systems and the like. I grew up playing with compass and straightedge.

Notice that nothing I have mentioned is in any way about hyperbolic geometry. I have told you how to draw a circle through J that meets two other circles orthogonally, the one with center A passing through CGDF, and the other passing through CMD.

It is possible to read about the "chordal," but I made up the other line, so I thought I had better include an example with numbers. Let the original circle be in the $xy$ plane with center A at $(0,0)$ and radius 5, so points CGDF would all satisfy $x^2 + y^2 = 25.$ But then let the point J lie along the $x$-axis at $(1,0).$ In this case the line I am calling $\Lambda$ becomes the vertical line at $x=13.$ Two example centers on $\Lambda:$ first, with center at $(13,0),$ the circle with radius 12 passes through J, and meets $x^2 + y^2 = 25$ at the two points $(\frac{25}{13}, \; \pm \frac{60}{13} ),$ and it is a minor exercise to show that the radii meet orthogonally, that is, the two circles are orthogonal. Second, take the circle with center at $(13,16)$ and radius 20. The circle does pass through J, and meets $x^2 + y^2 = 25$ at the two points $(-3,4)$ and $(\frac{77}{17}, \; \frac{-36}{17} ).$ Same comment about orthogonal circles, easy enough to verify. Indeed, the whole thing, in coordinates, is an exercise in second-year calculus, using two-variable functions and gradients.

  • $\begingroup$ I believe your method is correct, but could you please explain why your circle pass though J that will perpendicular to the circle CGDF and CMD? Please give me some hint? $\endgroup$
    – van abel
    Oct 31, 2011 at 3:16
  • $\begingroup$ The Figure of Your methods $\endgroup$
    – van abel
    Oct 31, 2011 at 3:23
  • $\begingroup$ Well, H is the intersection of $\Lambda$ and $\Delta.$ And H is the center of the final arc, FJMG. Because H is on $\Lambda,$ the arc is orthogonal to CDGF. Because H is on $\Delta,$ the arc is orthogonal to CMD if and only if it is orthogonal to CDGF. So, in fact, the arc FJMG is orthogonal to both. Please give it a try with actual compass and straightedge, it is not difficult to draw. $\endgroup$
    – Will Jagy
    Oct 31, 2011 at 3:24
  • $\begingroup$ your figure "The Figure of Your methods" is correct, assuming that things that seem to be right triangles really are so, and the distance AJ is the same as the distance XY, and so on. Have you taken multivariable calculus? Do you know how to find the gradient of a function of two variables? $\endgroup$
    – Will Jagy
    Oct 31, 2011 at 3:35
  • $\begingroup$ Ok, in fact, my question about your method is that why the center of circle which are perpendicular to circle A is must on the line $\Lambda$. Ps. $\nabla f=\rm{grad}(f(x,y))=(f_x,f_y)$, the gradient of a multi-variable function is a vector. $\endgroup$
    – van abel
    Oct 31, 2011 at 16:50

A description of ruler-and-compass construction in the Poincaré disk model can be found in this paper. Using the "hyperbolic straightedge" and "hyperbolic compass" described in the paper, you can carry out the ordinary procedure for erecting perpendiculars from Euclidean geometry (which doesn't depend of the fifth postulate).

  • $\begingroup$ There is something wrong with the pdf of Goodman-Strauss. It seems fine on the screen. Printing, the first two pages come out alright (38,39) then page 40 is moved all the way to the left paper edge and the diagrams are splintered. $\endgroup$
    – Will Jagy
    Oct 31, 2011 at 19:03
  • $\begingroup$ Must be a problem with your pdf-printing program, then. Perhaps try a different one (xpdf versus ghostscript). $\endgroup$ Oct 31, 2011 at 19:06

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