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…

Question

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

Aicharya · Accepted Answer

$\begin{aligned} & \because \frac{16+\alpha}{3}=6 \text { and } \frac{15+\beta}{3}=6 \\ & \Rightarrow(\alpha, \beta) \equiv(2,3)\end{aligned}$ Also, $\mathrm{C}_1 \mathrm{C}_2=\mathrm{r}_1+\mathrm{r}_2$ $\begin{aligned} & \Rightarrow \sqrt{(2-8)^2+\left(3-\frac{15}{2}\right)^2}=2 r_2+r_2 \\ & \Rightarrow r_2=\frac{5}{2} \Rightarrow r_1=2 r_2=5 \\ & \therefore(\alpha+\beta)+4\left(r_1^2+r_2^2\right) \\ & =5+4\left(\frac{25}{4}+25\right)=130\end{aligned}$

Let the circles $C_1:(x-\alpha)^2+(y-\beta)^2=r_1^2$ and $C_2:(x-8)^2+\left(y-\frac{15}{2}\right)^2=r_2^2$…

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