Find one moebius transformation which returns a segment for a given arc of a circle

Question

I'm doing some 2d geometry and I think the moebius transformation (as documented here https://en.wikipedia.org/wiki/M%C3%B6bius_transformation) can give me the answer I'm looking for, but I'm stuck.

I start from an arc of a circle, defined by its center, its radius, and two angles (at which the arc start and stop).

• Any transformation which can give map each point of the arc to a point of the segment would be good enough (bijection seems trivial here).
• A better solution would preserve the fact that M is at the middle of AB (so M' at the middle of A'B')
• The perfect solution (but I don't know if it exists) would be that the curvilinear abscissa of M on the arc is equal the the one of M' on the segment for all M in [A, B].

The position or the orientation of the resulting segment is not important.

What would be the best way to find the a, b, c, d parameters of the moebius transformation ?

Answer 1

$$\begin{bmatrix} B - M & AM-AB\\ A + B - 2 M & BM + AM - 2AB \end{bmatrix}$$ This sends any three points distinct points $A,B,M$ to $0,1,1/2$, respectively. Since mobius transformations take arcs/lines to arcs/lines and your three points are on a common arc, this will take that arc to the line from 0 to 1.

I computed this in Sage - you can replicate it by using the "point_transformation_matrix" method for projective space. There is probably an easier way, as to use that method you will have to projectivize your coordinates and then dehomogenize at the end.

Answer 2

HINT.- (1) The function $f$ (blue color in the attached figure) defined by $f(P)=Q$ is a bijection of the arc $AB$ on the line segment $CD$ contained in the tangent to the circle at the midpoint $M$.

(2) Apply this segment $CD$ on segment $A'B'$ using a function $g$ similar to $f$.

(3) One function that answers the problem is $g\circ f$.

Answer 3

Thank you @Piquito and @TokenToucan. Your answers are perfect, I couldn't get better, only more complex. My solution also miss the linearity of the coordinates after and before the projection (but it does use the möbius transformation :D)

I wanted to use the moebius formula, so while I waited for clues, I tried to find $a$, $b$, $c$ and $d$ of

$$ f(z) = \frac{a z + b}{c z + d} $$

Such as the three point of the arc $P_i$, $P_j$ and $P_k$ (respectively at start, middle and end of the arc) maps to $Q_i$, $Q_j$ and $Q_k$ (respectively at start, middle and end of a segment). And the result is hellish ! But seems to be correct.

$d$ can be fixed freely,

$$ d = 1 $$

$$ c = \frac{d \left(P_{i} Q_{j} - P_{i} Q_{k} - P_{j} Q_{i} + P_{j} Q_{k} + P_{k} Q_{i} - P_{k} Q_{j}\right)}{P_{i} P_{j} Q_{i} - P_{i} P_{j} Q_{j} - P_{i} P_{k} Q_{i} + P_{i} P_{k} Q_{k} + P_{j} P_{k} Q_{j} - P_{j} P_{k} Q_{k}} $$

$$ b = \frac{- P_{i} P_{j} Q_{i} c + P_{i} P_{j} Q_{j} c + P_{i} Q_{j} d - P_{j} Q_{i} d}{P_{i} - P_{j}} $$

$$ a = \frac{P_{i} Q_{i} c + Q_{i} d - b}{P_{i}} $$

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Which gives, for the example of @Piquito:

$$ P_i = (1, 0), P_j = \left( \frac{\sqrt 2}{2}, \frac{\sqrt 2}{2} \right ), P_k = (0, 1) $$

$$ Q_i = (\sqrt 2, 0), Q_j = \left( \frac{\sqrt 2}{2}, \frac{\sqrt 2}{2} \right ), Q_k = (0, \sqrt 2) $$

The following parameters for the möbius transformation :

$$ a = 2 + \sqrt 2$$ $$ b = -1 - i $$ $$ c = \frac{\sqrt 2}{2} - \frac{\sqrt 2}{2} i $$ $$ d = 1 $$

What is funny is that the result is clearly different from a geometric projection... but the curvilinear abscissa transformation is it not linear neither.