Let the circles C_1:(x- )^2+(y- )^2=r_2^1 and C_2:(x-8)^2+(y-(15)/(2))^2=r_2^2 touch each …

Question

Let the circles ${C}_{1}:(x-\alpha {)}^{2}+(y-\beta {)}^{2}={r}_{2}^{1}$ and ${C}_{2}:(x-8{)}^{2}+{(y-\frac{15}{2})}^{2}={r}_{2}^{2}$ 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 $(\alpha +\beta )+4({r}_{2}^{1}+{r}_{2}^{2})$ equals

Accepted Answer

$$\begin{aligned}
& C_1 \rightarrow(x-\alpha)^2+(y-\beta)^2=r_1^2 \
& C_2 \rightarrow(x-8)^2+\left(y-\frac{15}{2}\right)^2=r_2^2
\end{aligned}$$

Point $P$ divide the line segment internally $C_1 C_2$ in the ratio 2 : 1

$$\begin{aligned}
& \frac{\alpha \times 1+8 \times 2}{1+2}=6, \alpha=2 \
& \frac{\beta \times 1+\frac{15}{\alpha} \times 2}{1+2}=6, \beta=3 \
& r_1=\sqrt{(6-2)^2+(6-3)^2}=\sqrt{25}=5 \
& r_2=\sqrt{(8-6)^2+\left(\frac{15}{\alpha}-6\right)^2}=\frac{5}{2} \
& \alpha+\beta+4\left(r_1^2+r_2^2\right)=2+3+4\left(5^2+\left(\frac{5}{2}\right)^2\right) \
& =5+4\left(\frac{125}{4}\right) \
& =130 \
\end{aligned}$$

Solutions · 1