Suppose $M$ and $N$ are two hyperbolic lines in $(\mathbb{D},\mathcal{H})$ that don't intersect (hyperbolic geometry).
How do I find the hyperbolic line $L$ that is perpendicular to both $M$ and $N$?
I know that we can find the perpendicular line for just $M$, by having a point $z$ that isn't on $M$ and reflecting $z$ over $M$, $z^*$ then draw the hyperbolic line, $L$, through $z$ and $z^*$
How would I do this for both $M$ and $N$?
My other answer to your question here is fairly algebraic. I have a more geometric description for you as well here.
Assume your two given hyperbolic lines “end” in the ideal points $A$ and $B$ for line $M$ and the ideal points $C$ and $D$ for line $N$. Now intersect the Euclidean lines $AB$ and $CD$. (Let's for now assume they are not parallel, which we need to discuss in a later paragraph.) They intersect in a point $E$ which is the center of the circle that represents the line you are after.
Now you need a circle around that center and orthogonal to the unit circle. Construct the midpoint between that and $O$, the center of the unit disk, to get $F$. Draw a circle around $F$ through $O$ and $E$. It intersects the unit circle in two points, $G$ and $H$ (but see a later paragraph for cases when it doesn't). The circle around $E$ through $G$ and $H$ is the hyperbolic line you want.
If your two lines $AB$ and $CD$ are parallel, then the center of the circle you want is “at infinity” in a specific direction, namely the direction indicated by the parallel lines. In this case the line you want is the diameter of the unit circle, perpendicular to these two parallel lines.
If your two hyperbolic lines are not ultra-parallel (or hyper-parallel as Lee Mosher calls it), you won't get two distinct points $G$ and $H$. For limit-parallel lines you might get $E=G=H$ all sitting in the same ideal point on the unit circle, or for lines that intersect you would get $E$ inside the unit disk and thus the circle around $F$ not intersecting the unit circle at all.
As a comment by Blue already indicated, in the Poincaré disk model a hyperboloic line is modelled by a circle orthogonal to the unit circle, and angles are represented faithfully, so a hyperboloic line perpendicular to two given lines is modelled by a circle perpendicular to two given circles and the unit circle. All of these statements would include a straight line as a special case of a circle.
So how do you find a circle perpendicular to three given circles? Möbius geometry can help here, at least if you're happy to compute this instead of constructing this using geometric primitive operations. (Note my other answer for a more geometric approach.) A (non-oriented) circle is described by three numbers $(x,y,x^2+y^2-r^2)$. Turn this into a homogeneous coordinate vector
$$[x:y:x^2+y^2-r^2:1]=:[a:b:c:d]$$
with scalar multiples identified. Two circles are orthogonal if they satisfy
$$2a_1a_2+2b_1b_2-c_1d_2-d_1c_2=0\;.$$
If you have the canonical representative, i.e. $d_1=d_2=1$ then the above is equivalent to
$$(x_1-x_2)^2+(y_1-y_2)^2=r_1^2+r_2^2$$
and you can see Pythagoras in there as an indication that you have a right-angled triangle formed by the two centers and a point of intersection. And since the equation is homogenous, it works for other multiples of the same vector as well.
You can model a point as a circle of radius zero. A point lies on a circle if the above condition for orthogonality is satisfied. The homogeneous representation allows you to also include lines. A vector $[a:b:c:0]$ with a zero in the last coordinate corresponds to the line $2ax+2by-c=0$ so you can bring any line into that form quite easily.
So now you have three such vectors representing three generalized circles, and want a fourth representing the orthogonal circle. You want
$$\begin{pmatrix} 2a_1&2b_1&-d_1&-c_1\\ 2a_2&2b_2&-d_2&-c_2\\ 2a_3&2b_3&-d_3&-c_3 \end{pmatrix}\cdot \begin{pmatrix}a_4\\b_4\\c_4\\d_4\end{pmatrix}= \begin{pmatrix}0\\0\\0\end{pmatrix}\;.$$
In general you will have a 1-dimensional space of possible solutions here, which corresponds to the different representatives, different scalar multiples, of the same homogeneous coordinate vector. Pick any one. One easy choice would be using the minors:
\begin{align*} a_4&=\begin{vmatrix} 2b_1&-d_1&-c_1\\ 2b_2&-d_2&-c_2\\ 2b_3&-d_3&-c_3 \end{vmatrix}& b_4&=-\begin{vmatrix} 2a_1&-d_1&-c_1\\ 2a_2&-d_2&-c_2\\ 2a_3&-d_3&-c_3 \end{vmatrix}\\[2ex] c_4&=\begin{vmatrix} 2a_1&2b_1&-c_1\\ 2a_2&2b_2&-c_2\\ 2a_3&2b_3&-c_3 \end{vmatrix}& d_4&=-\begin{vmatrix} 2a_1&2b_1&-d_1\\ 2a_2&2b_2&-d_2\\ 2a_3&2b_3&-d_3 \end{vmatrix} \end{align*}
Now if you get $d_4\neq0$ you can turn that into a circle with center $\left(\frac{a_4}{d_4},\frac{b_4}{d_4}\right)$ and radius $\sqrt{\frac{a_4^2+b_4^2-c_4d_4}{d_4^2}}$. Watch out for cases where that numerator for the radius becomes negative; those solutions wouldn't really correspond to any circles with real radius. If $d_4=0$ your solution circle is in fact a line $2a_4x+2b_4y-c_4=0$.
I'm going to expand my comment from your previous deleted question into a limited answer: in short, the hyperbolic line $L$ need not exist. So whatever method you use for finding $L$ (such as the method in the answer of @MvG), you had better first check whether $L$ even exists.
Let me state what is true --- a necessary and sufficient condition for existence of $L$. For this purpose I'll need some notations and definitions.
First, recall that a line $M$ in the hyperbolic plane is represented in one of two ways in the Poincare disc model $\mathbb D$: either as the interior of a diameter of the closed disc $\overline{\mathbb D} = \mathbb D \cup S^1$, that diameter thus meeting $S^1$ orthogonally at its two endpoints; or as the interior of a circular arc in $\overline{\mathbb D}$ that meets $S^1$ orthogonally at the two endpoints of that arc. In either case, the hyperbolic line $M$ has a two point "ideal boundary" on $S^1$, denoted $\partial M \subset S^1$, and consisting either of the endpoints of the diameter, or the endpoints of the circular arc.
Next, consider two different hyperbolic lines $M,N$ in the hyperbolic plane. Since they are different, it follows that their endpoint sets are not equal: $\partial M \ne \partial N$. There are two cases to consider, depending on whether the intersection of their endpoint sets $\partial M \cap \partial N$ consists of zero points or one point.
In the first case where $\partial M \cap \partial N = \emptyset$, we say that $M$ and $N$ are hyperparallel.
In the second case where $\partial M \cap \partial N$ is a single point $\xi \in S^1$, the terminology is less well established. That link above calls them horoparallel or limiting parallel. But I like to say that $M$ and $N$ are asymptotic for the reason that $M$ and $N$ come arbitrarily close to each other in the hyperbolic metric, as one heads out to infinity in the direction of $\xi$. The geometric appearance of this case is that the two Euclidean circular arcs in $\overline{\mathbb D}$ that determine $M$ and $N$ meet tangentially at one common endpoint $\xi = \partial M \cap \partial N$.
Theorem: Given disjoint hyperbolic lines $M,N$ in the hyperbolic plane, the following are equivalent:
- There exists a hyperbolic line $L$ that is perpendicular to $M$ and $N$.
- $M$ and $N$ are hyperparallel.
As for determining whether $L$ exists, whenever the lines $M$ and $N$ are given to you in a computationally feasible manner, you should start by calculating their ideal boundaries $\partial M$ and $\partial N$ (e.g. in the Euclidean plane, intersect their corresponding circles with $S^1$). From there, you should be able to determine using $\partial M$ and $\partial N$ whether or not $M$ and $N$ intersect, and whether or not they are hyperparallel, by simply inspecting the configurations in $S^1$ of the two point subsets $\partial M, \partial N \subset S^1$.
Only in the case that they are hyperparallel should you then go on to determine their common perpendicular (e.g. by the method of @MvG).
