Q88 / 160 · AP EAMCET 2022 (07 Jul Shift 2) · Mathematics
The ratio of the areas of the greatest and the smallest circles touching $(x \pm 1)^2+(y \pm 1)^2=1$ is
- A$\frac{\sqrt{3}+1}{\sqrt{3}-1}$
- B$\frac{3+\sqrt{2}}{3-\sqrt{2}}$
- Correct answerC$\frac{3+2 \sqrt{2}}{3-2 \sqrt{2}}$
- D$4$
Solution
(c) Let radius of smaller circle be $r$ and radius of bigger.circle be $R$

$\begin{aligned} r & =\sqrt{2}-1 \\ R & =\sqrt{2}+1 \\ \therefore \text { Ratio of areas } & =\frac{(\sqrt{2}+1)^2}{(\sqrt{2}-1)^2}=\frac{3+2 \sqrt{2}}{3-2 \sqrt{2}}\end{aligned}$

$\begin{aligned} r & =\sqrt{2}-1 \\ R & =\sqrt{2}+1 \\ \therefore \text { Ratio of areas } & =\frac{(\sqrt{2}+1)^2}{(\sqrt{2}-1)^2}=\frac{3+2 \sqrt{2}}{3-2 \sqrt{2}}\end{aligned}$
Hence, the correct answer is option: 'C'.