Figure 6: Construct the hyperbolic line for the case in which
points A and B are not collinear with O and at
least one of the points is in the interior of γ. Begin by determining
the inverses of A and B with respect to γ. If A is
in the interior of γ, then by Theorem 2 the circle c2 passing
through points A, A’ and B is orthogonal to γ. If
A is on γ then A and A’ coincide (points on the
circle of inversion are invariant) and c2 can be constructed
using points A, B and B’. Determine the
intersections P2 and Q2 of c2 and γ and use them as
the endpoints of circular arc a2. Locate an additional point
M2 on arc a2. If A is on γ, choose M2 =
B , otherwise choose M2 = A. Finally construct arc
a2 using points P2, M2 and Q2.
Prepare for this construction by hiding all of the objects from the
previous construction except for points A and B and the
defining circle γ. Then construct the circular arc using the GeoGebra
code provided below. Note the use of the Reflect[
<Object>, <Circle> ]
command to determine the inverses of A and B with respect
to γ and the Length[ <Object> ] command to
determine whether or not A is on γ. The If[
<Condition>, <Then>,
<Else> ] command is used to construct the
circle passing through points A and B and orthogonal to γ.
If A is on γ, then the circle is constructed using A,
B and B’, otherwise it is constructed using points
A , A’ and B.