Calling the radius of the top left circle ‘r’, and the missing length we seek ‘x’, we can work our way around the rectangle, and express each of the radii in terms of r and x. We can form two right angles triangles and form their Pythagorean equations as below:

(1) x^2+(3-r)^2=(3+x-3r)^2

(2) x^2+r^2=(6-x-3r)^2

Taking the difference of these two, the x^2 terms cancel:

(3-r)^2-r^2=(3+x-3r)^2-(6-x-3r)^2

Factoring each side as a difference of squares:

(3-2r)(3)=(3+x-3r-6+x+3r)(3+x-3r+6-x-3r)

(9-6r)=(-3+2x)(9-6r)

If (9-6r)=0, r=3/2, but that would mean the top right circle would disappear entirely, so that’s not possible.

Since (9-6r) is not equal to 0, we can divide both sides by it:

1=(-3+2x)

4=2x

x=2

And that’s the answer!