JEE Math Practice 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

1. A.130
2. B.110
3. C.145
4. D.125