Copyright © University of Cambridge. All rights reserved.
'Centre Square' printed from https://nrich.maths.org/
Let $r$ be the radius of each of the larger circles.
The sides of the square are equal to $r+1$, the sum of the two
radii.
The diagonal of the square is $2r$.
By Pythagoras, $$(r+1)^2+(r+1)^2 =
(2r)^2$$Simplifying gives: $$
2(r+1)^2 = 4 r^2$$ i.e.
$$(r+1)^2 =
2r^2$$
so$$r+1 = \sqrt{2}r$$
[$-\sqrt{2}r$ is not possible since $r+1> 0$].
Therefore $(\sqrt{2} - 1)r = 1$.
Hence $r= \frac {1}{\sqrt{2}-1} = \sqrt{2} + 1$.