Key Concepts and Formulas

• Circles Touching Coordinate Axes: A circle touching both axes has its center at $(\pm r, \pm r)$, where $r$ is the radius. The signs depend on the quadrant.
• Distance Formula: The distance between points $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
• External Tangency: If two circles with centers $C_1$ and $C_2$ and radii $r_1$ and $r_2$ touch externally, then $C_1C_2 = r_1 + r_2$.
• Section Formula: If point $P(x,y)$ divides the line segment joining $C_1(x_1, y_1)$ and $C_2(x_2, y_2)$ in the ratio $m:n$, then $P = \left( \frac{n x_1 + m x_2}{m + n}, \frac{n y_1 + m y_2}{m + n} \right)$.

Step-by-Step Solution

Step 1: Determine the center and radius of circle $C_1$

We are given that $C_1$ lies in the third quadrant, has a radius of 3, and touches both coordinate axes. Since the circle is in the third quadrant and touches both axes, its center must be at $(-3, -3)$. The radius is given as 3.

• Radius of $C_1$: $r_1 = 3$
• Center of $C_1$: $C_1 = (-3, -3)$

Step 2: Determine the center of circle $C_2$ and let its radius be $r_2$

We are given that the center of $C_2$ is $(1, 3)$. Let the radius of $C_2$ be $r_2$.

• Center of $C_2$: $C_2 = (1, 3)$
• Radius of $C_2$: $r_2$ (unknown)

Step 3: Use the external tangency condition to find $r_2$

Since $C_1$ and $C_2$ touch externally, the distance between their centers equals the sum of their radii: $C_1C_2 = r_1 + r_2$.

• Calculate the distance $C_1C_2$: $C_1C_2 = \sqrt{(1 - (-3))^2 + (3 - (-3))^2} = \sqrt{(1+3)^2 + (3+3)^2} = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52}$
• Apply the external tangency condition: $\sqrt{52} = 3 + r_2$
• Solve for $r_2$: $r_2 = \sqrt{52} - 3$

Step 4: Find the point of tangency $(\alpha, \beta)$ using the section formula

The point of tangency $(\alpha, \beta)$ divides the line segment joining the centers $C_1$ and $C_2$ in the ratio of their radii $r_1:r_2 = 3:(\sqrt{52} - 3)$. Applying the section formula:

• $\alpha = \frac{r_2 x_1 + r_1 x_2}{r_1 + r_2} = \frac{(\sqrt{52} - 3)(-3) + (3)(1)}{3 + (\sqrt{52} - 3)} = \frac{-3\sqrt{52} + 9 + 3}{\sqrt{52}} = \frac{12 - 3\sqrt{52}}{\sqrt{52}}$
• $\beta = \frac{r_2 y_1 + r_1 y_2}{r_1 + r_2} = \frac{(\sqrt{52} - 3)(-3) + (3)(3)}{3 + (\sqrt{52} - 3)} = \frac{-3\sqrt{52} + 9 + 9}{\sqrt{52}} = \frac{18 - 3\sqrt{52}}{\sqrt{52}}$

Step 5: Calculate $(\beta - \alpha)^2$

• $\beta - \alpha = \frac{18 - 3\sqrt{52}}{\sqrt{52}} - \frac{12 - 3\sqrt{52}}{\sqrt{52}} = \frac{18 - 3\sqrt{52} - 12 + 3\sqrt{52}}{\sqrt{52}} = \frac{6}{\sqrt{52}}$
• $(\beta - \alpha)^2 = \left(\frac{6}{\sqrt{52}}\right)^2 = \frac{36}{52}$

Step 6: Simplify the fraction and find $m+n$

• Simplify the fraction $\frac{36}{52}$ by dividing both numerator and denominator by their greatest common divisor, 4: $\frac{36}{52} = \frac{9}{13}$
• Identify $m$ and $n$: $m = 9$ and $n = 13$. Since $\operatorname{gcd}(9, 13) = 1$, the condition is satisfied.
• Calculate $m+n$: $m + n = 9 + 13 = 22$

Common Mistakes & Tips

• Sign Errors: Be careful with signs when dealing with circles in different quadrants. The center coordinates are negative in the third quadrant.
• Simplifying Radicals: Simplifying radicals like $\sqrt{52}$ can help avoid errors in later calculations. $\sqrt{52} = 2\sqrt{13}$.
• Section Formula Ratio: Ensure the correct order of radii in the section formula.

Summary

We found the centers and radii of the two circles using the given information and the condition for external tangency. Then, we used the section formula to find the point of tangency $(\alpha, \beta)$. Finally, we calculated $(\beta - \alpha)^2$, simplified the result to $\frac{9}{13}$, and found $m+n = 22$.

Final Answer

The final answer is \boxed{22}, which corresponds to option (A).