Question 319088
{{{drawing(300,300,-10,10,-10,10,
circle(-2,2,2.9),circle(2,-2,2.9),
circle(-2,2,0.2),circle(2,-2,0.2),
locate(-1.5,4,R),
locate(2.5,-3,R),
locate(-1,2,R),
locate(0.5,-1,R),
green(line(-2,2,-2,-2)),
green(line(-2,-2,2,-2)),
blue(line(-2,-2,-2,-5)),
blue(line(-2,2,-2,5)),
blue(line(-2,2,2,-2)),
blue(line(2,-2,2,-5)),
circle(-5,5,0.3),
circle(-5,-5,0.3),
circle(5,-5,0.3),
circle(5,5,0.3),
line(-5,5,-5,-5),
line(-5,-5,5,-5),
line(5,-5,5,5),
line(5,5,-5,5))}}}
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I think this is what you mean.
You can find the length of the side of the square by using the radii of the circle.

The length of the green line can be calculated using the hypotenuse {{{2R}}} since that triangle is a right isoceles triangle. 
The green line length, {{{L}}} is,
{{{L^2+L^2=(2R)^2}}}
{{{2L^2=4R^2}}}
{{{L=2R^2}}}
{{{L=sqrt(2)*R}}}
The side of the square is then,
{{{S=R+L+R}}}
{{{S=2R+sqrt(2)R}}}
{{{S=R(2+sqrt(2))}}}
So then the area of the square is,
{{{As=R^2(2+sqrt(2))^2}}}
and the area of the two circles is,
{{{Ac=2*pi*R^2}}}
So then the area between the two would be the difference,
{{{highlight(A=R^2((2+sqrt(2))^2-2*pi)^2)) }}}or approximately,
{{{A=5.37R^2}}}
For R=1,
{{{A=5.37}}}