Theorem HA6
Suppose that K,L are hyperbolic lines, and that XεC.
There is a horocycle at X touching K,L if and only if
X lies on an axis for K,L.
proof
Suppose that J is a horocycle at X which touches K at U and L at V.
Let H be the axis of the hyperbolic lines XU, XV, and let h denote inversion in H.
As a hyperbolic line, H is orthogonal to C, and hence to J as J,C touch at X.
Thus, h(J) = J. Also, h swaps the hyperbolic lines XU,XV. Since U,V are intersections
of these with J, h swaps U,V. As K,L are the perpendiculars to XU,XV at U,V,
h swaps K,L and hence lies on an axis for these hyperbolic lines.
Conversely, suppose X lies on H, an axis for K,L, and that h denotes inversion in H.
By the tangency lemma, there is a horocycle J at X touching K at U, the foot of the
perpendicular from X to K. As X is on H, h fixes X. As H is an axis, h swaps K,L.
It follows that h maps U to V, the foot of the perpendicular from X to L.
By the argument in the first paragraph, h(J) = J, so J touches L at V.
Thus, J is a horocycle of the required type.
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