Yes.
A "7-hexagon" is formed by joining 7 circles as follows:
[asy][asy]draw(Circle((0,0),1));
draw(Circle((2,0),1));
draw(Circle((-2,0),1));
draw(Circle((1,sqrt(3)),1));
draw(Circle((-1,-sqrt(3)),1));
draw(Circle((-1,sqrt(3)),1));
draw(Circle((1,-sqrt(3)),1));
[/asy][/asy]
A "10-hexagon" is formed by joining 10 circles as follows:
[asy][asy]draw(Circle((0,0),1));
draw(Circle((2,0),1));
draw(Circle((-2,0),1));
draw(Circle((1,sqrt(3)),1));
draw(Circle((-1,-sqrt(3)),1));
draw(Circle((-1,sqrt(3)),1));
draw(Circle((1,-sqrt(3)),1));
draw(Circle((3,sqrt(3)),1));
draw(Circle((3,-sqrt(3)),1));
draw(Circle((4,0),1));
[/asy][/asy]
Now, to place the 2012 circles, one only needs to place 286 pieces of 7-hexagons and one 10-hexagon anywhere in the plane.
Generalized version:
For which positive integers $n$ is it possible to place $n$ distinct circles with the same diameter on the plane, such that each circle touches at least three others circles?