Square, rhombus and a circle

As shown in the following image you have a rhombus of maximum area inside a unit square. Find radius of the circle.
https://skydrive.live.com/embed?cid=8779…

I'm not sure exactly what is meant by a "rhombus of maximum area" here, but I suppose solving a sort of general form of the problem may help.

Suppose that the rhombus has sides length x < 1.
We find length GC = 1 - x
The length FC may be take to be, by the Pythagorean theorem, √(x^2 - (1 - x)^2) = √(2x - 1).
The length DF is then given by 1 - FC = 1 - √(2x - 1)

As the figure is drawn, one may traverse the diameter of the circle by beginning at the upper-left corner, going down a length of FC = 1 - √(2x - 1) downward, then going a length of GC = 1 - x across. Thus, we may solve for the radius of the circle by stating

(2r)^2 = 4r^2 = (1 - √(2x - 1))^2 + (1 - x)^2
= 1 - 2√(2x - 1) + 2x - 1 + x^2 - 2x + 1
= x^2 - 2√(2x - 1) + 1

Again, I'm not entirely sure about the terms of the question, but this seems useful

at the lowest possible length of x = 1/2, the radius reaches its maximum possible value of
4r^2 = 1/4 + 1 = 5/4
r^2 = 5/16
r = √5/4

EDIT:
are you sure that FD = CG?
if FD = CG, then
1 - FD = 1 - CG
CF = BG
CF = GF
which means GFC is an isosceles triangle with 2 90˚ angles, which is possible if(f) the rhombus is the same as the square containing it, which means that r = 0.

I was thinking on falzoon's last question and started thinking on this line. Obviously did not think in the correct direction.

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Get some sleep :)

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The size of rhombus = 1-x
So (1-x)^2=x^2+h^2
The area of rhombus =x. h
A=( 1-2x)^1/2. .x
dA/dx=1/2 (x^2-2x^3)^-1/2. (2x-6x^2)= 0
Awill max if x= 1/3. And h =1/3V3
The rhombus area is = 2/3. 1/3V3=2/9V3
So the size of rhombus = 2/3
The radius of the circle inside rhombus = area:(2x)
r=2/9V3:(2.1/3)=1/3V3

I'm confused.
Wouldn't the rhombus with maximum area inside the unit square just be the unit square?
That would make the circle a point (or if you like a curcle with radius 0).