Let $A B C D$ and $A E F G$ be squares of side 4 and 2 units, respectively. The point $E$ is on the line segment $\mathrm{AB}$ and the point $\mathrm{F}$ is on the diagonal $\mathrm{AC}$. Then the radius $\mathrm{r}$ of the circle passing through the point $\mathrm{F}$ and touching the line segments $\mathrm{BC}$ and $\mathrm{CD}$ satisfies:

• A

  $r=0$
• B

  $2 r^2-4 r+1=0$
• C

  $2 r^2-8 r+7=0$
• Correct answer
  D

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

Solution

$\begin{aligned} & \mathrm{OF}^2=\mathrm{r}^2 \\ & (2-r)^2+(2-r)^2=\mathrm{r}^2 \\ & \mathrm{r}^2-8 \mathrm{r}+8=0\end{aligned}$

Hence, the correct answer is option: 'D'.