(Here, $\mathbb{H}^2$ is the hyperbolic plane.)

Let $\vec{u}=\left(-\frac35, \frac45\right)$, $\vec{b} = \left(\frac65, \frac25\right)$, $s=\sqrt{\|\vec{b}\|^2-1}$. Let $C$ be the Eucledian circle in $\mathbb{R}^2$ centred at $\vec{b}$ of radius $s$, and $L=C\cap \mathbb{H}^2$. Find the centre $\vec{a}\in\mathbb{R}^2$ and the radius $r$ of the Eucledian circle $D$ in $\mathbb{R}^2$ such that $M=D\cap \mathbb{H}^2$ is the hyperbolic line which is asymptotic to $\vec{u}$ and intersects orthogonally with $L$.

Here's my sketch of the problem:

enter image description here

I have come up with two equations:

$$\textbf{(1) } \|\vec{a}-\vec{u}\|^2 = r^2$$ $$\textbf{(2) } \text{Orthogonality condition for two circles}$$

But I have a hard time figuring out what a third equation could be. Would appreciate some advice.

  • $\begingroup$ Not sure what so hyperbolic is about your question. Better formulate everything first in hyperbolic geometry terms. Then what model you use and after that the euclidean construction I know it sounds much more work but would improve our and your own understanding greatly (and that is worth it :) $\endgroup$ – Willemien Jul 23 '17 at 9:23
  • $\begingroup$ @Willemien I think I formulated it in terms of hyperbolic geometry. I used the Poincaré disc model, as taught. $\endgroup$ – sequence Jul 23 '17 at 19:38

I suppose you are working with the Poincare disk model ( please check in Wikipedia)

And you have to find the details (let's call them that) of the hyperbolic line which is asymptotic to  $u$ and intersects orthogonal with $L$

The hyperbolic line that you are looking for is not only orthogonal to $L$ it must also be orthogonal to the unit circle (otherwise it is just no hyperbolic line)

The euclidean centre of the euclidean circle you are looking for is on the intersection of:

  • the euclidean line tangent to the unit circle going through $u$
  • the euclidean line going to (both) intersections of the circle $C$ and the unit circle .

With the above you can calculate point $ a $ and also its radius .

Good luck

Ps there is a mistake in your drawing the centre $a$ is outside the unit disk

  • $\begingroup$ I took from your message the hint that the Eucledian circle related to the hyperbolic line $M$ is orthogonal to the Poincaré disc, which I dind't notice eventually. Thank you for that. I didn't, however, understand the rest of your message (or why it is true). So what I did is that I now got the third equation for another orthogonality condition and solved the three equations to get $\vec{a}=\left(-\frac{17}{9},-\frac{1}{6}\right)$, $r=\frac{29}{18}$. This is very tedious, but I don't know how to do any better. Geometry is not my thing. $\endgroup$ – sequence Jul 23 '17 at 20:24
  • $\begingroup$ No not good A is around (1/3 , 3/2) (not exact) I do it by construction not by Algebra , all centres of circles that are orthogonal to the unit circle at u are on the tangent at u the other line is the line containing the centres of all circles that are orthogonal to circle C and the unit circle so the centre of the only circle having all those properties is on the intersection $\endgroup$ – Willemien Jul 23 '17 at 21:15
  • $\begingroup$ Also the circle C you did draw is much to big (it doesn't go through (0,0) maybe best to make a new sketch $\endgroup$ – Willemien Jul 23 '17 at 21:23
  • $\begingroup$ Unless I made some inattention error (typo) in my equations, the values should be correct. $\endgroup$ – sequence Jul 23 '17 at 21:34
  • 1
    $\begingroup$ A circle with radius 2.65 doesn't even intersect the unit circle $\endgroup$ – Willemien Jul 23 '17 at 22:07

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