Question

Let $R$ be the region of the disc $x^2+y^2\le 1$ in the first quadrant. Then the area of the largest possible circle contained in $R$ is

• A. $\pi (3-2\sqrt{2})$
• B. $\pi (4-3\sqrt{2})$
• C. $\frac{\pi }{6}$
• D. $\pi (2\sqrt{2}-2)$

Answer 1

The correct option is A $\pi (3-2\sqrt{2})$

We know that
$OA=\sqrt{r^2+r^2}=r\sqrt{2}$
Let $OB=R=1$ then
$R=r+r\sqrt{2}\Rightarrow r=\frac{1}{1+\sqrt{2}}\Rightarrow r=\sqrt{2}-1$

Hence the area of the inscribed circle will be
$=\pi (\sqrt{2}-1{)}^2=\pi (2+1-2\sqrt{2})=\pi (3-2\sqrt{2})$