Algebraic solutions for Poincaré Disk arcs

Question

Given two points on the Poincaré Disk, there is a single straight line or arc that passes through them and that is orthogonal to the unit circle. Using compass and straightedge methods, one can easily plot the circle that generates that arc (shown in blue below).

Is there a simple algebraic method that can be used to express the radius and location of the generating circle in terms of the two points in the complex plane? I have tried writing down the constraints, but the process becomes rather ugly.

Any ideas?

N.B: Geometric Construction:

In the diagram, the black circle is the unit circle / Poincaré Disk, and the red points are the points between which the blue arc should be drawn.

1. Draw a line from the center through one of the red points.
2. Draw a perpendicular line to the first through the same point.
3. Draw the tangent lines to the unit circle at the intersection points.
4. The tangents intersection is the third point which is needed to define the blue circle, which is a hyperbolic line.

Answer 1

For further reference:

If the two points are $P$ and $Q$, then the arc center $(x_0, y_0)$ is given by:

$$x_0 = \frac{ q_y (p_x^2 + p_y^2 + 1) - p_y (q_x^2 + q_y^2 + 1)}{2(p_x q_y - p_y q_x)}$$ $$y_0 = \frac{-q_x (p_x^2 + p_y^2 + 1) + p_x (q_x^2 + q_y^2 + 1)}{2(p_x q_y - p_y q_x)}$$ And the arc radius is given by $r_0 = \sqrt{x_0^2 + y_0^2 - 1}$. I have not fond an elegant way to present this without dissecting the complex coordinate into it's parts.

Answer 2

I've figured, from the formulas in nbubis answer, if we let $p = p_x + p_y i$ and $q = q_x + q_y i$, then the formulas for the center of the circle and its radius can also be written as:

$$c = \frac{p(q\bar q + 1) - q(p\bar p + 1)}{p\bar q - q\bar p} \qquad r = \sqrt{c\bar c - 1}\;.$$