The ratio of the areas of the greatest and th | vedclass

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(C) The given circles are $(x-1)^2+(y-1)^2=1$,$(x+1)^2+(y-1)^2=1$,$(x-1)^2+(y+1)^2=1$,and $(x+1)^2+(y+1)^2=1$. These circles have radius $1$ and cente

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$\frac{3+2\sqrt{2}}{3-2\sqrt{2}}$

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(C) The given circles are $(x-1)^2+(y-1)^2=1$,$(x+1)^2+(y-1)^2=1$,$(x-1)^2+(y+1)^2=1$,and $(x+1)^2+(y+1)^2=1$. These circles have radius $1$ and centers at $(\pm 1, \pm 1)$.The distance of the centers from the origin $(0,0)$ is $\sqrt{1^2+1^2} = \sqrt{2}$.The smallest circle is centered at the origin and touches these four circles. Its radius $r$ is the distance from the origin to the center of any of the given circles minus the radius of those circles: $r = \sqrt{2} - 1$.The greatest circle is centered at the origin and encloses these four circles. Its radius $R$ is the distance from the origin to the center of any of the given circles plus the radius of those circles: $R = \sqrt{2} + 1$.The ratio of the areas is $\frac{\pi R^2}{\pi r^2} = \frac{(\sqrt{2}+1)^2}{(\sqrt{2}-1)^2} = \frac{2+1+2\sqrt{2}}{2+1-2\sqrt{2}} = \frac{3+2\sqrt{2}}{3-2\sqrt{2}}$.

List-$I$	List-$II$
$A$. $x^2+y^2+2x+8y-23=0$,$x^2+y^2-4x-10y+19=0$	$I$. $0$
$B$. $x^2+y^2=1$,$x^2+y^2-2x-6y+6=0$	$II$. $1$
$C$. $x^2+y^2-8x+2y=0$,$x^2+y^2-2x-16y+25=0$	$III$. $2$
$D$. $x^2+y^2=4$,$x^2+y^2-2x=0$	$IV$. $3$
	$V$. $4$

The ratio of the areas of the greatest and the smallest circles touching $(x \pm 1)^2 + (y \pm 1)^2 = 1$ is

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