jmc

Construct the square $ABCD$ by connecting the centers of the large circles, as shown, and consider the isosceles right $△BAD$.



Since $AB=AD=2r$ and $BD=2+2r$, we have $2(2r{)}^2=(2+2r{)}^2$. So $$1+2r+r^2=2r^2,\text{and}{r}^2-2r-1=0.$$Applying the quadratic formula gives $r=\boxed{1+\sqrt{2}}$.