Let n geq 3 and let C_1, C_2, ldots, C_n be c | vedclass

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(D) Let the angle between the $X$-axis and the line $y=2 \sqrt{2} x+10$ be $2 	heta$. The slope of the line is $m = 	an(2 	heta) = 2 \sqrt{2}$.Usin

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a geometric progression with common ratio $2+\sqrt{3}$

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(D) Let the angle between the $X$-axis and the line $y=2 \sqrt{2} x+10$ be $2 	heta$. The slope of the line is $m = 	an(2 	heta) = 2 \sqrt{2}$.Using the formula $	an(2 	heta) = \frac{2 	an 	heta}{1-	an^2 	heta}$,we have $\frac{2 	an 	heta}{1-	an^2 	heta} = 2 \sqrt{2}$,which simplifies to $\sqrt{2} 	an^2 	heta + 	an 	heta - \sqrt{2} = 0$.Solving for $	an 	heta$,we get $(\sqrt{2} 	an 	heta - 1)(	an 	heta + \sqrt{2}) = 0$. Since $	heta$ is acute,$	an 	heta = \frac{1}{\sqrt{2}}$.Then $\sin 	heta = \frac{1}{\sqrt{3}}$.Let $P$ be the intersection of the two lines. For any circle $C_i$ with radius $r_i$ and center $O_i$,the distance from $P$ to $O_i$ is $d_i = \frac{r_i}{\sin 	heta} = \sqrt{3} r_i$.Since the circles touch externally,the distance between centers $O_i$ and $O_{i+1}$ is $r_i + r_{i+1}$. Also,$d_{i+1} = d_i + r_i + r_{i+1}$.Substituting $d_i = \sqrt{3} r_i$,we get $\sqrt{3} r_{i+1} = \sqrt{3} r_i + r_i + r_{i+1}$,which simplifies to $r_{i+1}(\sqrt{3}-1) = r_i(\sqrt{3}+1)$.Thus,$\frac{r_{i+1}}{r_i} = \frac{\sqrt{3}+1}{\sqrt{3}-1} = \frac{(\sqrt{3}+1)^2}{3-1} = \frac{4+2 \sqrt{3}}{2} = 2+\sqrt{3}$.Therefore,the radii form a geometric progression with common ratio $2+\sqrt{3}$.

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