Is it consistent that there is a model of $\mathsf{ZFC}$ (or $\mathsf{ZF}$) with the following properties:


(1) For all $x \in {}^\omega 2$, $x^\sharp$ exists (or $\mathbf{\Sigma}\_1^1$ determinacy)


(2) For all sets $A$, there exists $x \in {}^\omega 2$ such that $A \in L[x]$, i.e. every set is constructible from a real. 


If one takes an arbitrary model $V$ of $\mathsf{ZFC}$ satisfying (1), will $L[({}^\omega2)^V]$ have this property or does it necessarily add a set that is not constructible from any single real? 


Thanks for any information or clarification.



Sharps and Every Set is Constructible from a Real
The theory is not consistent, since $\mathbb{R}$ is a set, but if $\mathbb{R}\in L[x]$, then every real is in $L[x]$, and this contradicts the existence of $x^\sharp$. 


