We present some results on orthogonal ilines. These will have important applications
to hyperbolic geometry.
Theorem O1
Suppose that A and B are distinct points, and that L is an iline.
If A, B are not inverse with respect to L, then there is a unique
iline through A and B orthogonal to L.
If A, B are inverse with respect to L, then any iline 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 iline 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 iline.
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 ilines. Recall that the apollonian
family A(A,B) consists of all ilines 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 ilines, 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 ilines through
the points A and B.
If A is a point on the iline L, then T(A,L) denotes the family of ilines
which touch L at A.
If L and M are ilines, then O(L,M) is the family of ilines 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 ilines, that A,B are distinct points,
and that K,N are orthogonal ilines 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 iline through A is an extended line.
Observe that, if E is an extended line, then any iline 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 ilines 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 ilines 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).

