hyperbolic geometry (and circle ) construction problem Was thinking about hyperbolic geometry, the Poincare Disk Model and Sweikarts constant and combined them all in a construction puzzle that I was unable to solve.
My construction puzzle:
Given:


*

*A circle $Circle_0$ with centre $Centre_0$ and radius $r$

*On $Circle_0$ we have 2 points $I_1$ and $I_2$

*Trough point $I_1$ orthogonal (perpendicular)  to $Circle_0$ is circle $Circle_1$

*Trough point $I_2$ orthogonal (perpendicular) to $Circle_0$ is circle $Circle_2$

*$Circle_1$ and $Circle_2$ have the same radius

*$Circle_1$ and $Circle_2$ are orthogonal to eachother.

*point Q is the point inside $Circle_0$ where $Circle_1$ and $Circle_2$  cut eachother.


Wanted: construct point Q
the only limits I could find are:


*

*Q is on the line perpendiculer to $ I_1I_2$ going to the midpoint of $ I_1I_2$

*Q is on the same site as side of $Centre_0$ as   $ I_1$ and $I_2$

*$ \angle   I_1Centre_0I_2$ is smaller than a right angle


I did manage the opposite:
Given point Q (different from $Centre_0$ ) construct the points $I_1$ and $I_2$
so if it helps somebody:


*

*Draw ray $r$ from $Centre_0$ trough Q

*Draw line l trough Q perpendicular to ray r

*Point $ I_c$ where line l cuts $Circle_0$ (any of the two)

*Draw segment $Circle_0$ to Point $ I_c$

*Draw line $j$ trough $ I_c$ perpendicular to the segment$Circle_0$ $ I_c$

*Point $ I_Q$ where line $j$  cuts ray $r$

*Point $ I_m$ is the midpoint of the segment $Q$ $I_Q$

*Line $m$ trough $ I_m$ perpendicular to ray $r$

*Draw $Circle_m$ centre $ I_m$ trough Q

*Point $ Centre_1$ where line $m$  cuts $ Circle_m$ (one of the two)

*Point $ Centre_2$ where line $m$  cuts $ Circle_m$ (the other one)

*Draw $Circle_1$ centre $ Centre_1$ and trough Q 

*Draw $Circle_2$ centre $ Centre_2$ and trough Q

*Point $I_1$ is where $Circle_1$ cuts $Circle_0$ nearest to Q

*Point $I_2$ is where $Circle_2$ cuts $Circle_0$ nearest to Q 


But now from $ Circle_0 $ , $I_1$ and $I_1$ how can I construct $Point Q$ ?
 A: Not an answer, but too long for a comment.

Let your $Circle_0$ be the unit circle centered at the origin, and let $I_1$ and $I_2$ be mutual reflections in the $x$-axis. Take $J_i$ to be the center of a circle through $I_i$ orthogonal to the unit circle; then $J_i$ lives on the line tangent to the circle at $I_i$. The intersections, say, $P$ and $Q$, of $\bigcirc J_1$ and $\bigcirc J_2$ (each of sufficiently-large radius, $r$) live on the $x$-axis. If the circles are to be orthogonal at these points, then their tangents at $P$ and $Q$ make $45^\circ$ angles with the $x$ axis; thus, $\overline{PQ}$ is necessarily one side of an (the) axis-aligned square inscribed in $\bigcirc J_1$ (and, likewise, of the corresponding square in $\bigcirc J_2$). This says that each $J_i$ must be at distance $r/\sqrt{2}$ from the $x$-axis.
We may assume the coordinates of $I_1$ are $(\cos\theta, \sin\theta)$ for some $0 \leq \theta \leq \pi/2$; and the coordinates of $J_1$ become $(\cos\theta \pm r\sin\theta, \sin\theta \mp r \cos\theta)$, where the $\pm$ and $\mp$ signs depend upon which direction $\overrightarrow{I_1J_1}$ points along the tangent line at $I_1$. The distance from $J_1$ is therefore $\sin\theta \mp r \cos\theta$. By the above, we must have
$$\sin\theta \mp r \cos\theta = r/\sqrt{2} \qquad \to \qquad r = \frac{\sqrt{2}\sin\theta}{1\pm\sqrt{2}\cos\theta}$$
To construct $Q$ (and $P$), then, all you need to do is construct a radius of length $r$. Radius $r$ is certainly constructible, but I don't (yet?) have a construction that fits nicely with the rest of the figure.
