Let $$ABCD$$ and $$AEFG$$ be squares of side 4 and 2 units, respectively. The point $$E$$ is on the line segment $$AB$$ and the point $$F$$ is on the diagonal $$AC$$. Then the radius $$r$$ of the circle passing through the point $$F$$ and touching the line segments $$BC$$ and $$CD$$ satisfies:

Let $$ABCD$$ be a square of side 4 and $$AEFG$$ be a square of side 2, where $$E$$ is on $$AB$$ and $$F$$ is on diagonal $$AC$$.

Place $$A=(0,0)$$, $$B=(4,0)$$, $$C=(4,4)$$, and $$D=(0,4)$$. Since $$E$$ is on $$AB$$ with $$AE=2$$, it follows that $$E=(2,0)$$, and the diagonal $$AC$$ has equation $$y=x$$.

Square $$AEFG$$ of side 2 has vertices in order: starting from $$A=(0,0)$$ and $$E=(2,0)$$, going counterclockwise gives $$F=(2,2)$$ and $$G=(0,2)$$. One can verify that $$EF=FG=GA=AE=2$$ and that $$F(2,2)$$ lies on $$y=x$$.

We seek a circle through $$F$$ tangent to lines $$BC: x=4$$ and $$CD: y=4$$. Since these lines are perpendicular, the center must be equidistant from both, so it is $$(4-r,4-r)$$ with radius $$r$$.

Imposing that $$F(2,2)$$ lies on the circle gives

$$ (2 - (4 - r))^2 + (2 - (4 - r))^2 = r^2 $$

$$ 2(r - 2)^2 = r^2 $$

$$ 2r^2 - 8r + 8 = r^2 $$

$$ r^2 - 8r + 8 = 0 $$

The answer is Option D: $$r^2 - 8r + 8 = 0$$.