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Let the circles $C_1:(x-\alpha)^2+(y-\beta)^2=r_1^2$ and $C_2:(x-8)^2+\left(y-\frac{15}{2}\right)^2=r_2^2$ touch each other externally at the point $(6,6)$. If the point $(6,6)$ divides the line segment joining the centres of the circles $C_1$ and $C_2$ internally in the ratio $2: 1$, then $(\alpha+\beta)+4\left(r_1^2+r_2^2\right)$ equals
$110$
$130$
$125$
$145$
Solution
$ \because \frac{16+\alpha}{3}=6 \text { and } \frac{15+\beta}{3}=6 $
$ \Rightarrow(\alpha, \beta) \equiv(2,3) $
$ \text { Also, } \mathrm{C}_1 \mathrm{C}_2=\mathrm{r}_1+\mathrm{r}_2 $
$ \Rightarrow \sqrt{(2-8)^2+\left(3-\frac{15}{2}\right)^2}=2 \mathrm{r}_2+\mathrm{r}_2 $
$ \Rightarrow \mathrm{r}_2=\frac{5}{2} \Rightarrow \mathrm{r}_1=2 \mathrm{r}_2=5 $
$ \therefore(\alpha+\beta)+4\left(\mathrm{r}_1^2+\mathrm{r}_2^2\right) $
$ =5+4\left(\frac{25}{4}+25\right)=130$
