Let R be the region of the disc x^2+y^2 leq 1 | vedclass

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(A) Let the radius of the largest circle contained in the first quadrant region $R$ be $r$. The center of this circle will be at $(r, r)$ because it m

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

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(A) Let the radius of the largest circle contained in the first quadrant region $R$ be $r$. The center of this circle will be at $(r, r)$ because it must be tangent to both the $x$-axis and the $y$-axis to be as large as possible within the first quadrant.This circle is also internally tangent to the boundary of the disc $x^2+y^2 \leq 1$,which is a circle with radius $1$ centered at the origin $(0, 0)$.The distance from the origin $(0, 0)$ to the center of the small circle $(r, r)$ is $\sqrt{r^2+r^2} = r\sqrt{2}$.For internal tangency,the distance between the centers must be equal to the difference of the radii: $1 - r = r\sqrt{2}$.Rearranging the terms,we get $1 = r(1+\sqrt{2})$,so $r = \frac{1}{\sqrt{2}+1} = \sqrt{2}-1$.The area of this circle is $\pi r^2 = \pi(\sqrt{2}-1)^2 = \pi(2 + 1 - 2\sqrt{2}) = \pi(3-2\sqrt{2})$.

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

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