The diagonals of a square intersect at right angles. The radius of the circle is a/2. If ABCD is the inscribed square and the centre of the circle is O, then AO=BO=CO=DO=a/2 and ∠AOB=90°.
Therefore AB=BC=CD=AD=√(a²/4+a²/4)=a/√2 or a√2/2.
The circle inscribed in square ABCD has diameter=AB, so its radius is AB/2=a√2/4.
The sum of the radii is a/2+a√2/4=0.8536 approx.
This can be written as a(2+√2)/4.