Key Concepts and Formulas

• Equation of a Circle: A circle with center $(h,k)$ and radius $r$ has the equation $(x-h)^2 + (y-k)^2 = r^2$.
• Circles Touching Externally: When two circles touch each other externally, their centers and the point of tangency are collinear, and the distance between their centers is equal to the sum of their radii ($C_1C_2 = r_1 + r_2$).
• Section Formula (Internal Division): If a point $P(x,y)$ divides the line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ internally in the ratio $m:n$, then the coordinates of $P$ are given by: $x = \frac{mx_2 + nx_1}{m+n}, \quad y = \frac{my_2 + ny_1}{m+n}$
• Distance Formula: The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.

Step-by-Step Solution

Step 1: Identify the Centers and the Point of Tangency

The given equations of the circles are:

• $C_1: (x-\alpha)^2+(y-\beta)^2=r_1^2$ From this equation, the center of circle $C_1$ is $C_1(\alpha, \beta)$ and its radius is $r_1$.
• $C_2: (x-8)^2+\left(y-\frac{15}{2}\right)^2=r_2^2$ From this equation, the center of circle $C_2$ is $C_2\left(8, \frac{15}{2}\right)$ and its radius is $r_2$.

We are given that the circles touch externally at the point $P(6,6)$.

Step 2: Use the Section Formula to Find $\alpha$ and $\beta$

• Explanation: Since the circles touch externally at $P(6,6)$, this point of tangency $P$ must lie on the line segment connecting the centers $C_1$ and $C_2$. The problem states that $P(6,6)$ divides the line segment $C_1C_2$ internally in the ratio $2:1$. This means that $C_1P:PC_2 = 2:1$.
• Let $C_1 = (x_1, y_1) = (\alpha, \beta)$ and $C_2 = (x_2, y_2) = \left(8, \frac{15}{2}\right)$.
• Let $P = (x,y) = (6,6)$.
• The ratio of division is $m:n = 2:1$, so $C_1P:PC_2 = 2:1$.

Applying the section formula for the x-coordinate: $x = \frac{m x_2 + n x_1}{m+n}$ $6 = \frac{2 \cdot 8 + 1 \cdot \alpha}{2+1}$ $6 = \frac{16 + \alpha}{3}$ $18 = \alpha + 16$ $\alpha = 18 - 16$ $\boxed{\alpha = 2}$

Applying the section formula for the y-coordinate: $y = \frac{m y_2 + n y_1}{m+n}$ $6 = \frac{2 \cdot \frac{15}{2} + 1 \cdot \beta}{2+1}$ $6 = \frac{15 + \beta}{3}$ $18 = \beta + 15$ $\beta = 18 - 15$ $\boxed{\beta = 3}$

So, the center of circle $C_1$ is $C_1(2,3)$.

Step 3: Calculate Radii $r_1$ and $r_2$

• Explanation: The point of tangency $P(6,6)$ lies on both circles. Therefore, the distance from the center of a circle to $P$ is equal to its radius.

For $r_1$: $r_1$ is the distance between $C_1(2,3)$ and $P(6,6)$. $r_1 = \sqrt{(6-2)^2 + (6-3)^2}$ $r_1 = \sqrt{(4)^2 + (3)^2}$ $r_1 = \sqrt{16 + 9}$ $r_1 = \sqrt{25}$ $\boxed{r_1 = 5}$ Thus, $r_1^2 = 25$.

For $r_2$: $r_2$ is the distance between $C_2\left(8, \frac{15}{2}\right)$ and $P(6,6)$. $r_2 = \sqrt{(8-6)^2 + \left(\frac{15}{2}-6\right)^2}$ $r_2 = \sqrt{(2)^2 + \left(\frac{15-12}{2}\right)^2}$ $r_2 = \sqrt{4 + \left(\frac{3}{2}\right)^2}$ $r_2 = \sqrt{4 + \frac{9}{4}}$ $r_2 = \sqrt{\frac{16+9}{4}}$ $r_2 = \sqrt{\frac{25}{4}}$ $\boxed{r_2 = \frac{5}{2}}$ Thus, $r_2^2 = \left(\frac{5}{2}\right)^2 = \frac{25}{4}$.

Step 4: Evaluate the Final Expression

We need to find the value of $(\alpha+\beta)+4\left(r_1^2+r_2^2\right)$. Substitute the values we found: $\alpha=2$, $\beta=3$, $r_1^2=25$, and $r_2^2=\frac{25}{4}$.

$(\alpha+\beta)+4\left(r_1^2+r_2^2\right) = (2+3) + 4\left(25 + \frac{25}{4}\right)$ $= 5 + 4\left(\frac{100}{4} + \frac{25}{4}\right)$ $= 5 + 4\left(\frac{125}{4}\right)$ $= 5 + 125$ $= \boxed{130}$

Common Mistakes & Tips

• Ratio Confusion: Ensure the correct order of points and ratios in the section formula. $C_1P:PC_2 = 2:1$.
• Calculation Errors: Double-check arithmetic, especially when squaring fractions and adding terms.
• Understanding External Tangency: Remember that the point of tangency lies on the line segment connecting the centers of two externally tangent circles.

Summary

This problem requires applying the section formula and distance formula in the context of externally tangent circles. By finding the centers and radii of the circles, the desired expression can be evaluated. The final answer is 130.

The final answer is \boxed{130}, which corresponds to option (A).