The construction is based on a well ordering of $R^3$ into the least ordinal of cardinality continuum. Let $\phi$ be that ordinal and let $R^3={p_\alpha:\alpha<\phi}$ R^3=\{p_\alpha:\alpha<\phi\}$ be an enumeration of the points of space. We define a unit circle$C_\alpha$containing$p_\alpha$by transfinite recursion on$\alpha$, for some$\alpha$we do nothing. Here is the recursion step. Assume we have reached step$\alpha$and some circles ${C_\beta:\beta<\alpha}$\{C_\beta:\beta<\alpha\}$ have been determined. If some of them contains (=covers) $p_\alpha$, p_\alpha$, we do nothing. Otherwise, we choose a unit circle containing $p_\alpha$ that misses all the earlier circles. For that, we first choose a plane throu $p_\alpha$ that is distinct from the planes of the earlier circles. This is possible, as there are continuum many planes throu $p_\alpha$ and less than continuum many planes which are the planes of those earlier circles. Let$K$be the plane chosen. The earlier circles intersect$K$in less than continuum many points, so it suffices to find, in$K$, a unit circle going throu $p_\alpha$ which misses certain less than continuum many points. That is easy: there are continuum many unit circles in$K$taht pass throu $p_\alpha$ and each of the bad points disqualifies only 2 of them. 1 The construction is based on a well ordering of$R^3$into the least ordinal of cardinality continuum. Let$\phi$be that ordinal and let$R^3={p_\alpha:\alpha<\phi}$be an enumeration of the points of space. We define a unit circle$C_\alpha$containing$p_\alpha$by transfinite recursion on$\alpha$, for some$\alpha$we do nothing. Here is the recursion step. Assume we have reached step$\alpha$and some circles${C_\beta:\beta<\alpha}$have been determined. If some of them contains (=covers)$p_\alpha$, we do nothing. Otherwise, we choose a unit circle containing$p_\alpha$that misses all the earlier circles. For that, we first choose a plane throu$p_\alpha$that is distinct from the planes of the earlier circles. This is possible, as there are continuum many planes throu$p_\alpha$and less than continuum many planes which are the planes of those earlier circles. Let$K$be the plane chosen. The earlier circles intersect$K$in less than continuum many points, so it suffices to find, in$K$, a unit circle going throu$p_\alpha$which misses certain less than continuum many points. That is easy: there are continuum many unit circles in$K$taht pass throu$p_\alpha\$ and each of the bad points disqualifies only 2 of them.