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.