Let the circles C_1: (x-alpha)^2 + (y-beta)^2 | vedclass

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(B) Let the centers be $C_1(\alpha, \beta)$ and $C_2(8, \frac{15}{2})$. The point of contact $P(6,6)$ divides $C_1C_2$ in the ratio $r_1:r_2$. Since t

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$130$

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(B) Let the centers be $C_1(\alpha, \beta)$ and $C_2(8, \frac{15}{2})$. The point of contact $P(6,6)$ divides $C_1C_2$ in the ratio $r_1:r_2$. Since the circles touch externally,the distance between centers is $C_1C_2 = r_1 + r_2$. Given that $P(6,6)$ divides $C_1C_2$ in the ratio $2:1$,we have $r_1:r_2 = 2:1$,so $r_1 = 2r_2$. Using the section formula for point $P(6,6)$ dividing $C_1C_2$ in ratio $2:1$: $6 = \frac{2(8) + 1(\alpha)}{2+1}$ $\Rightarrow 18 = 16 + \alpha$ $\Rightarrow \alpha = 2$. $6 = \frac{2(\frac{15}{2}) + 1(\beta)}{2+1}$ $\Rightarrow 18 = 15 + \beta$ $\Rightarrow \beta = 3$. Now,$C_1C_2 = \sqrt{(8-2)^2 + (\frac{15}{2}-3)^2} = \sqrt{6^2 + (\frac{9}{2})^2} = \sqrt{36 + \frac{81}{4}} = \sqrt{\frac{144+81}{4}} = \sqrt{\frac{225}{4}} = \frac{15}{2}$. Since $C_1C_2 = r_1 + r_2 = 2r_2 + r_2 = 3r_2$,we have $3r_2 = \frac{15}{2} \Rightarrow r_2 = \frac{5}{2}$ and $r_1 = 5$. Finally,$(\alpha+\beta) + 4(r_1^2 + r_2^2) = (2+3) + 4(5^2 + (\frac{5}{2})^2) = 5 + 4(25 + \frac{25}{4}) = 5 + 100 + 25 = 130$.

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