Sibling Circles

By pythagorean theorem, we have

$(R+1)^2=(2-R)^2+(1-R)^2$

$R^2+2R+1=4-4R+r^2+1-2R+R^2$

$R^2-8R+4=0$

Solving for $R$ using the quadratic formula, we have

$R=\dfrac{8 \pm \sqrt{(-8)^2-4(4)}}{2}=4\pm \dfrac{\sqrt{48}}{2}=4 \pm 2\sqrt{3}$

$R \approx 0.53589$ or $R \approx 7.4641$

The desired answer is $\boxed{R \approx 0.53589}$ since $R \approx 7.4641$ is greater than the side length of the square.