Golfing Flower of Life
Mathematica, 177 173 128 124 120 bytes

c=Circle;Graphics@{{0,0}~c~3,Rotate[Table[If[-3&lt;x-y&lt;4,c[{√3x,-x+2y}/2,1,Pi/{6,2}]],{x,-3,2},{y,-4,2}],Pi/3#]&amp;~Array~6}




The main idea is to compose the result from six rotated versions of this:



This in turn is a rectangular table of identical circle arcs with two corners cut off. If we remove the shearing and represent each circle center with a #, we basically want to distribute the circles in this pattern:

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These edges are cut off by imposing the condition -3 &lt; x-y &lt; 4 on the 2D indices (since the value of x-y is constant along diagonals) and the shearing comes from multiplying these x and y by non-orthogonal basis vectors which span the grid we're looking for.

This particular orientation of the unrotated arcs turns out to be the shortest since both ends of the arc evenly divide Pi so that the arc can be expressed as Pi/{6,2} (all other arcs would either require and additional minus sign or integers in the numerator).