By a Tarski plane I understand a mathematical structure $(X,B,\equiv)$ consisting of set $X$, a betweenness relation $B\subseteq X^3$ and a congruence relation ${\equiv}\subseteq X^2\times X^2$ satisfying Tarski's axioms minus (Axiom schema of Continuity and Euclid axiom). So, this structure satisfies 9 axioms.
A Tarski plane is called (non-)Euclidean if it satisfies (the negation of) the Euclid axiom.
To fulfill standard constructions with ruler and compass on a Tarski plane, one needs to accept two continuity axioms:
Segment-Circle Axiom: $\forall c,r,x,y,p,q\in X$ $(Bcxr\wedge Bcry\wedge cp\equiv cx\wedge cq\equiv cy)\to\exists z\in X (Bpzq\wedge cz\equiv cr)$
and
Circle-Circle Axiom: $\forall a,c,r,x,y,p,q\in X$ $(Bcxr\wedge Bcry\wedge cp\equiv cx\wedge cq\equiv cy\wedge ap\equiv aq)\to\exists z\in X (az\equiv ap\wedge cz\equiv cr)$.
Due to Strommer, we know that for Euclidean Tarski planes, the Segment-Circle Axiom is equivalent to the Circle-Circle Axiom. Moreover, the argument of Theorem 11.6 in this book seems to prove that the Circle-Circle Axiom implies the Segment-Circle Axiom in any Tarski plane.
Problem. Does the Segment-Circle Axiom imply the Circle-Circle Axiom in every Tarski plane?
Remark. In Remark 11.6.1 Hartshorne writes that the Segment-Circle Axiom and Circle-Circle Axiom are equivalent in an arbitrary Hilbert plane but the proof of this fact is not direct and follows from the classification theorem of Pejas. But it is not clear if this classification theorem of Pejas is applicable to any Tarski plane.
