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

Answer: (A)

$\pi (3-2\sqrt{2})$

Let the radius of the inscribed circle be $r$.

$OP=OC+CP$

Using Pythagoras Theorem, $OC=r\sqrt{2}$.

$\Rightarrow$ $r\sqrt{2}+r=1$

$\Rightarrow$ $r=\frac{1}{\sqrt{2}+1}$

$\Rightarrow$ $r=\sqrt{2}-1$

Area of circle $=$ $\pi r^2$

$=\pi (\sqrt{2}-1{)}^2$

$=\pi (3-2\sqrt{2})$

Therefore, option A is correct.