We present some results on orthogonal i-lines. These will have important applications to hyperbolic geometry. Theorem O1 Suppose that A and B are distinct points, and that L is an i-line. If A, B are not inverse with respect to L, then there is a unique i-line through A and B orthogonal to L. If A, B are inverse with respect to L, then any i-line through A and B is orthogonal to L. proof If A ≠ ∞, we invert the figure in a circle with centre A, and denote the images by A' = ∞, B' and L'. Any i-line through A' must be an extended line. Also, A,B are inverse with respect to L if and only if A',B' are inverse with respect to L'. If L' is an extended line, then A'εL', so A', B' are not inverse with respect to L'. There is a unique extended line through B' orthogonal to L'. Inverting this again, we get the required i-line. Now suppose that L' is a circle, with centre C. Note that A' and C are inverse with respect to L'. If A and B are not inverse with respect to L, then B' ≠ C, so there is a unique extended line through B' orthogonal to L', namely the line B'C. If A and B are inverse with respect to L, then B' = C, so that any extended line through B' is orthogonal to L'. In either case, inverting again gives the relevant result. The next result concerns families of i-lines. Recall that the apollonian family A(A,B) consists of all i-lines for which A, B are inverse points. We know that the members of the family are disjoint. In fact, there is a converse in the Common Inverses Theorem which states that, for any pair L,M of disjoint i-lines, there is a unique pair {A,B} such that L,M belong to A(A,B). definitions If A,B are distinct points, then B(A,B) is the family of i-lines through the points A and B. If A is a point on the i-line L, then T(A,L) denotes the family of i-lines which touch L at A. If L and M are i-lines, then O(L,M) is the family of i-lines orthogonal to both L and M. It is easy to see that if g is an inversive transformation, then g(A(A,B)) = A(g(A),g(B)), g(B(A,B)) = B(g(A),g(B)), g(T(A,L)) = T(g(A),g(L)), and g(O(L,M)) = O(g(L),g(M)). Theorem O2 Suppose that L and M are distinct i-lines, that A,B are distinct points, and that K,N are orthogonal i-lines through A. If L,MεB(A,B), then O(L,M) = A(A,B). If L,MεT(A,K), then O(L,M) = T(A,N). If L,MεA(A,B), then O(L,M) = B(A,B). proof If A ≠ ∞, we can invert in a circle centre A. Thus, we may assume that A = ∞ so any i-line through A is an extended line. Observe that, if E is an extended line, then any i-line orthogonal to E is an extended line perpendicular to E or a circle with its centre on E. In the first case, L,M are lines through B. No extended line can be perpendicular to both, so i-lines orthogonal to both mst be circles with centre B. Thus O(L,M) = A(∞,B) = A(A,B). In the second case, L,M are parallel lines. Now no circle can have its centre on both, so i-lines orthogonal to both are the extended lines perpendicular to both, and hence parallel. Thus O(L,M) = T(A,N). In the third case, L,M must be circles with centre B (as A = ∞). Suppose that GεO(L,M) and that g denotes inversion in G. As G is orthogonal to L,M, g(L) = L and g(M) = M, so that g(A(A,B)) = A(A,B). It folows that g maps {A,B} to {A,B}. Since g maps the interior of L to itself, so g fixes A and B. Thus A,B lie on G, i.e. GεB(A,B).
Theorem O3 Suppose that K,N are orthogonal and belong to O(L,M). Then L,M are disjoint, and N is uniquely determined by K. proof If L,M intersect, then we have one of the first two cases in Theorem O2. But then O(L,M) consists of disjoint i-lines (the first case) or touching i-lines (the second case). In neither case can the family contain a pair of orthogonal i-lines. It follows that we must have L,MεA(A,B) for some A,B. By inverting if necessary, we may assume that A = ∞, so that the i-lines orthogonal to L,M are the extended lines through B. Thus, given K, N is uniquely determined.

main asymptotic page