We will suppose that ΔABC is such that F = F(cosh(a),cosh(b),cosh(c)) > 0,
and write Δ = Δ(cosh(a),cosh(b),cosh(c)).
Then the circumradius equation
is FR^{2} = Δ, and, as F > 0, it has a positive solution R.
If the circumcircle exists, then its radius must satisfy cosh(r) = R.
Now, cosh(x) > 1 for x ≠ 0, so we will have no circumcircle if R ≤ 1.
In fact we prove something more powerful and useful.
size lemma With ΔABC as above, and R the positive solution of the equation, we have R = cosh(r), where 2r ≥ max{a,b,c}.
proof
Our object is a sufficiency proof, but we note that the lemma indicates


The lemma brings us close to our sufficiency proof. Assuming that F > 0, we have a solutrion of the circumradius equation R = cosh(r). Consider the hyperbolic circles K(A,r), K(B,r) and K(C,r), where K(P,r) denotes the hyperbolic circle with hyperbolic centre P and hyperbolic radius r. By the size lemma, 2r ≥ a, so K(B,r) and K(C,r) do intersect at least once. In general they will meet twice, once on each side of the hyperbolic line AB. Similarly the other pairs meet. Thus we have up to six points of intersection. Now note that K(P,r) will be the circumcircle if and only if each of A,B,C is on K(P,r). This means that P must lie on the three circles K(A,r), K(B,r), K(C,r). We need to show that one of the intersections of K(A,r) and K(B,r) is actually on K(C,r).
sufficient condition for a hyperbolic circumcircle
proof
By the basic lemma, t is such that
Now, r arises from the circumradius equation, which is really

