In the entire solution, when I say circle, I mean geometric circle.
With center (x,y,z) = (1/2,0,0), place circles in the xy-plane of radii .5, 1.5, 2.5, ..., i.e. at radius n+0.5, where n is nonnegative integer.
Notice that a sphere of any radius centered at (0,0,0) either intersects one of the above circles at two points, or has a point of tangency to each of two of the above circles. Remove from each sphere centered at (0,0,0) the two points that are already on circles. These punctured spheres need to be covered with additional circles.
Here is how to cover each of the above twice-punctured spheres with disjoint circles. Place tangent planes to the two removed points. If the planes are parallel, the two points are like the north and south poles of a globe, and you can cover the rest of the sphere with circles of latitude. If the tangent planes are not parallel, they intersect in a line. Call this line the "hinge line." The intersection of the sphere with any plane containing this hinge line is a circle (as long as it is non-empty). In addition, since a line and a point not on the line determine a plane, every point on the sphere is on a unique plane containing the hinge line. Thus this covers each twice-punctured sphere by disjoint circles.
The disjoint union of the circles on the xy-plane with center (1/2,0,0) and radius n+0.5 along with the above twice-punctured spheres is all of R^3. Thus all of R^3 is covered with disjoint geometric circles.
Thanks to Dan Asimov for telling me about this puzzle and solution. It is discussed in the following book:
Halmos, Paul R., "Problems for Mathematicians, Young and Old," Mathematical Association of America, 1991.
Here is another solution to the circle puzzle by Oliver Goodman (firstname.lastname@example.org): Take an open spherical ball. Remove a circle of half the radius going through the center and one pole. The intersection of a plane perpendicular to the removed circle and going through the pole is an open disk minus a point. Open disk minus a point can be made out of circles. Since all these open disks minus a point are disjoint we can make the open ball minus a circle out of circles. Putting back the circle we get an open ball plus a point. Join copies of these pole to pole along a line like a string of beads. So now we have exactly covered the union of a line and a collection of touching open balls. Fill the remaining space with circles centered on and perpendicular to the line.
Created: May 18 1995 --- Last modified: Jun 18 1996