Is it true that an algebraically closed field $k$ definable (in the model theory sense) in a real closed field $\mathcal R$ is the algebraic closure $\mathcal{R}^{alg}=\mathcal{R}(\sqrt{-1})$?

UPDATE: if we prove that the definable set $K \subset \mathcal{R}^n$ that defines $k$ in $\mathcal R$ has the same cardinality as $\mathcal R$ then by categoricity of $ACF_0$ we get that $K$ is isomorphic to $\mathcal{R}(\sqrt{-1})$ (which is definable in $\mathcal R$ in an obvious fashion). Would this isomorphism be definable?