Many problems in hyperbolic geometry can be solved using the Poincare disk model.
In particular it is often useful to apply a hyperbolic transformation so that one of the
points is at the centre of the disk. Then any hyperbolic line through this point is also
a euclidean diameter of the disk. We can relate the distances in each geometry.
some notation
Suppose that A,B lie in the disk. Then we define
s(AB) = sinh(½d(A,B))
c(AB) = cosh(½d(A,B))
t(AB) = tanh(½d(A,B))
AB = euclidean distance between A and B.
the hybrid distances theorem
Suppose that A and B are points in the disk, and O is the centre of the disk.
Then AB = s(AB)/c(OA)c(OB).
proof
Since we know more about the euclidean geometry of the disk, this is useful.
For example, we know that a hyperbolic circle is a euclidean circle which lies
entirely in the disk. We can characterize the hyperbolic triangles which have
hyperbolic circumcircles algebraically. This involves the quantity Δ which we
first met in the sine rule. This can be expressed in terms of s(AB), s(BC),
and s(CA), but this involves a root.
the hyperbolic circumcircle theorem
The hyperbolic triangle ΔABC
has a hyperbolic circumcircle if and only if
4s(AB)s(BC)s(CA) < Δ.
If the condition is satisfied, then the hyperbolic radius of the circumcircle
is given by r, where tanh(r) = 4s(AB)s(BC)s(CA)/Δ.
proof
Since a hyperbolic triangle has Δ > 0, we may restate the condition as
"if and only if Δ^{2}  (4s(AB)s(BC)s(CA))^{2} > 0".
If we now use the formula for Δ^{2}, and the identity cosh(2x) = 2sinh^{2}(x)+1,
we find (after some algebraic simplification) that
Δ^{2}  (4s(AB)s(BC)s(CA))^{2} = 4H(s(AB),s(BC),s(CA)),
where H(x,y,z) = (x+y+z)(x+yz)(y+zx)(z+xy).
The condition now becomes
"if and only if H(s(AB),s(BC),s(CA)) > 0."
We can also use identities to get a formula for the hyperbolic radius in terms
of H. Summing up, we get
the hyperbolic circumcircle theorem  version 2
The hyperbolic triangle ΔABC
has a hyperbolic circumcircle if and only if
H(s(AB),s(BC),s(CA)) > 0.
If the condition is satisfied, then the hyperbolic radius of the circumcircle
is r, where sinh^{2}(r) = 4s^{2}(AB)s^{2}(BC)s^{2}(CA)/H(s(AB),s(BC),s(CA)).
Note that the expression for sinh^{2}(r) is equal to (2R)^{2}, where R is the
radius of the euclidean circumcircle of the triangle with sides of length
s(AB),s(BC),s(CA). This triangle exists as H(s(AB),s(BC),s(CA)) > 0.
proof
Only the second part remains to be proved. It follows from the identity
sinh^{2}(x) = tanh^{2}(x)/(1tanh^{2}(x)), and earlier calculations.
The proof of the first version also yields
a result on euclidean circumcircles
If A,B,C lie in the disk, then the euclidean circumcircle of ABC is
(1) a circle within D if H(s(AB),s(BC),s(CA)) > 0,
(2) a circle touching C if H(s(AB),s(BC),s(CA)) = 0, and
(3) an iline cutting C twice if H(s(AB),s(BC),s(CA)) < 0.
As we already know, case (1) corresponds to a hyperbolic circle, and case (2)
to a horocycle. Later, we shall discuss the third category as hypercircles.
We shall refer to H as Heron's Polynomial, since it occurs in Heron's formula.
Here, we note merely that it also characterizes the triples (k,l,m) of positive
numbers for which there is a triangle with sides of length k,l,m. Such a
triangle exists if and only if the sum of any two exceeds the
third, and this is equivalent to H(k,l,m) > 0. We give a proof here. It works equally well in
euclidean and hyperbolic geometry.
If (k,l,m) satisfy the condition, we will refer to a corresponding triangle as
a (k,l,m) triangle. It is, of course, unique up to conjugacy.
We may now restate our result as
the hybrid circumcircle theorem
The hyperbolic triangle ΔABC
has a hyperbolic circumcircle if and only if
there exists a euclidean (s(AB),s(BC),s(CA)) triangle.
When the circumcircle exists, its hyperbolic radius r is given by
sinh(r) = 2R, where R is the euclidean radius of the second triangle.
We use the term hybrid to describe a theorem which relates hyperbolic
and euclidean properties.

