Let Φ be a fixed point free group given by the presentation \documentclass[12pt]{minimal}
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\begin{document}$$\langle A, B\,\vert\, A^\mu=1,\, B^\nu=A^t,\, BAB^{-1}=A^\rho\rangle$$\end{document} where μ and ρ are relative prime numbers, t = μ/s and s = gcd(ρ − 1,μ), and ν is the order of ρ modulo μ. We prove that if (1) ν = 2, and (2) Φ is embeddable into the multiplicative group of some skew field, then Φ is circular. This means that there is some additive group N on which Φ acts fixed point freely, and |(Φ(a)+b)∩(Φ(c)+d)| ≤ 2 whenever a,b,c,d ∈ N, a≠0≠c, are such that Φ(a)+b≠Φ(c)+d.