1. Identify the centres of the circles

For $C_1:(x-\alpha)^2+(y-\beta)^2=r_1^2,$ its centre is $O_1=(\alpha,\beta).$

For $C_2:(x-8)^2+\left(y-\frac{15}{2}\right)^2=r_2^2,$ its centre is $O_2=\left(8,\frac{15}{2}\right).$

The circles touch externally at $P=(6,6).$

2. Use the ratio condition

The point $P=(6,6)$ divides the line segment joining the centres $O_1$ and $O_2$ internally in the ratio $2:1$.

This means $O_1P:PO_2=2:1.$

Using section formula, if $P$ divides $O_1O_2$ internally in the ratio $2:1$, then $P=\left(\frac{2x_2+1x_1}{3},\frac{2y_2+1y_1}{3}\right)$ where $(x_1,y_1)=O_1=(\alpha,\beta)$ and $(x_2,y_2)=O_2=\left(8,\frac{15}{2}\right)$.

So, $6=\frac{2\cdot 8+\alpha}{3}$ $18=16+\alpha$ $\alpha=2.$

Also, $6=\frac{2\cdot \frac{15}{2}+\beta}{3}$ $18=15+\beta$ $\beta=3.$

Hence, $\alpha+\beta=2+3=5.$

3. Use the touching condition to find radii

Since the circles touch externally at $P$, the point $P$ lies on both circles, and the distances from centres to $P$ are the radii.

Thus, $r_1=O_1P,\qquad r_2=O_2P.$

For $C_1$:

Centre $O_1=(2,3)$, point $P=(6,6)$.

$r_1^2=(6-2)^2+(6-3)^2=4^2+3^2=16+9=25.$ So, $r_1=5.$

For $C_2$:

Centre $O_2=\left(8,\frac{15}{2}\right)$, point $P=(6,6)$.

$r_2^2=(6-8)^2+\left(6-\frac{15}{2}\right)^2$ $=(-2)^2+\left(-\frac{3}{2}\right)^2=4+\frac{9}{4}=\frac{25}{4}.$

So, $r_2=\frac{5}{2}.$

4. Compute the required expression

We need $(\alpha+\beta)+4(r_1^2+r_2^2).$

Substitute the values: $\alpha+\beta=5,$ $r_1^2=25,\qquad r_2^2=\frac{25}{4}.$

Then $r_1^2+r_2^2=25+\frac{25}{4}=\frac{100+25}{4}=\frac{125}{4}.$

Therefore, $4(r_1^2+r_2^2)=4\cdot \frac{125}{4}=125.$

Hence, $(\alpha+\beta)+4(r_1^2+r_2^2)=5+125=130.$

5. Check with options

The value is $130.$ So the correct option is A.

6. Compare with stored correct answer

Stored correct answer: A

Our derived answer: A

They agree.