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(A) The four circles are centered at $(\pm r, \pm r)$ with radius $r$. These circles are located in the four quadrants. Let the circle touching all fo

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$2\sqrt{2}$

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(A) The four circles are centered at $(\pm r, \pm r)$ with radius $r$. These circles are located in the four quadrants. Let the circle touching all four be centered at the origin $(0,0)$ with radius $R$. The distance from the origin to the center of any of the four circles is $\sqrt{r^2 + r^2} = r\sqrt{2}$. For the circle to touch these circles externally,$R + r = r\sqrt{2} \implies R = r(\sqrt{2}-1)$. For the circle to touch these circles internally,$R - r = r\sqrt{2} \implies R = r(\sqrt{2}+1)$. Thus,$r_1 = r(\sqrt{2}-1)$ and $r_2 = r(\sqrt{2}+1)$. Then,$r_1 + r_2 = r(\sqrt{2}-1 + \sqrt{2}+1) = 2r\sqrt{2}$. Therefore,$\frac{r_1+r_2}{r} = 2\sqrt{2}$.

If $r_1$ and $r_2$ are the radii of two circles that touch all four circles given by $(x \pm r)^2 + (y \pm r)^2 = r^2$,then find the value of $\frac{r_1+r_2}{r}$.

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