1. Set up coordinates for the bigger square

Let the square $ABCD$ have coordinates $A(0,0),\quad B(4,0),\quad C(4,4),\quad D(0,4).$ Then the diagonal $AC$ is the line $y=x.$

2. Place the smaller square $AEFG$

Since $AEFG$ is a square of side $2$, and $E$ lies on segment $AB$, we must have $E=(2,0).$ Now $F$ lies on diagonal $AC$, and $AEFG$ is a square built on side $AE$.

Starting from $A(0,0)$ and $E(2,0)$, the square can be taken inside $ABCD$, so $F=(2,2),\quad G=(0,2).$ Indeed, $F$ lies on $y=x$, so this matches the condition.

3. Equation of the required circle

The circle touches line segments $BC$ and $CD$.

• $BC$ is the line $x=4$
• $CD$ is the line $y=4$

A circle tangent to both these perpendicular lines must have its center at equal distance $r$ from each line. Hence the center is $(4-r,\,4-r).$

So the radius is $r$, and the circle passes through $F(2,2)$.

4. Use the condition that the circle passes through $F$

Distance from center $(4-r,4-r)$ to $F(2,2)$ must equal $r$: $\sqrt{(4-r-2)^2+(4-r-2)^2}=r.$ That is, $\sqrt{(2-r)^2+(2-r)^2}=r,$ $\sqrt{2(2-r)^2}=r.$ Now square both sides: $2(2-r)^2=r^2.$ Expand: $2(r^2-4r+4)=r^2,$ $2r^2-8r+8=r^2,$ $r^2-8r+8=0.$

5. Match with the options

This is exactly: $\boxed{r^2-8r+8=0}$ which is Option D.

6. Check the radius values

Solving, $r=\frac{8\pm\sqrt{64-32}}{2}=4\pm2\sqrt{2}.$ Since the center must lie inside the square near corner $C$, we need $r<4$, so the valid radius is $r=4-2\sqrt{2}.$ This indeed satisfies the quadratic in option D.

Therefore, the correct option is: $\boxed{\text{D}}$