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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 $\mathrm{R}$ is
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Verified Answer
The correct answer is:
$\pi(3-2 \sqrt{2})$
Required equation of circle $(x-h)^{2}+(y-h)^{2}=h^{2}$
Both circle touch internally
$$
\begin{array}{l}
\mathrm{C}_{1} \mathrm{C}_{2}=\left|\mathrm{r}_{1}-\mathrm{r}_{2}\right| \\
\sqrt{\mathrm{h}^{2}+\mathrm{h}^{2}}=|\mathrm{h}-1|
\end{array}
$$
Solve this $\mathrm{h}=\sqrt{2}-1$
Area $\pi(\sqrt{2}-1)^{2}=\pi(3-2 \sqrt{2})$
Both circle touch internally
$$
\begin{array}{l}
\mathrm{C}_{1} \mathrm{C}_{2}=\left|\mathrm{r}_{1}-\mathrm{r}_{2}\right| \\
\sqrt{\mathrm{h}^{2}+\mathrm{h}^{2}}=|\mathrm{h}-1|
\end{array}
$$
Solve this $\mathrm{h}=\sqrt{2}-1$
Area $\pi(\sqrt{2}-1)^{2}=\pi(3-2 \sqrt{2})$
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