Let $A B C D$ be a square of side length 1 , and $\Gamma$ a circle passing through $B$ and $C$, and touching $\mathrm{AD}$. The radius of $\Gamma$ is

• A

  $\frac{3}{8}$
• B

  $\frac{1}{2}$
• C

  $\frac{1}{\sqrt{2}}$
• Correct answer
  D

  $\frac{5}{8}$

Solution

$\mathrm{PC}=\mathrm{r}$ $\mathrm{PC}^{2}=\mathrm{r}^{2}$ $(\mathrm{r}-1)^{2}+\left(\frac{1}{2}-1\right)^{2}=r^{2}$ $1-2 \mathrm{r}+\frac{1}{4}=0$ $r=\frac{5}{8}$

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