1. Equation and center of $C_1$

Since $C_1$ is a circle of radius $3$ in the third quadrant touching both coordinate axes, its center must be at $(-3,-3).$ So, $C_1: (x+3)^2+(y+3)^2=9.$

2. Center of $C_2$ and condition of external tangency

The center of $C_2$ is given as $(1,3)$.

Distance between the centers: $d=\sqrt{(1+3)^2+(3+3)^2}=\sqrt{4^2+6^2}=\sqrt{52}=2\sqrt{13}.$

If the circles touch externally, then the point of contact lies on the line joining the centers and divides the segment joining the centers internally in the ratio of their radii.

Let radius of $C_2$ be $r_2$. Since they touch externally, $3+r_2=2\sqrt{13} \implies r_2=2\sqrt{13}-3.$

3. Point of contact $(\alpha,\beta)$

Let

• center of $C_1 = A(-3,-3)$ with radius $r_1=3$,
• center of $C_2 = B(1,3)$ with radius $r_2=2\sqrt{13}-3$.

For external tangency, the contact point $P(\alpha,\beta)$ lies on $AB$ and $AP:PB=r_1:r_2=3:(2\sqrt{13}-3).$

Using section formula, $P=\left(\frac{3\cdot 1+(2\sqrt{13}-3)(-3)}{3+(2\sqrt{13}-3)},\;\frac{3\cdot 3+(2\sqrt{13}-3)(-3)}{3+(2\sqrt{13}-3)}\right).$

But this can be simplified more easily by using the unit direction vector from $A$ to $B$.

Vector from $A$ to $B$ is $B-A=(4,6).$ Its magnitude is $2\sqrt{13}.$ So unit vector along $AB$ is $\left(\frac{2}{\sqrt{13}},\frac{3}{\sqrt{13}}\right).$

Since the contact point is at distance $3$ from $A$ toward $B$, $P=A+3\left(\frac{2}{\sqrt{13}},\frac{3}{\sqrt{13}}\right).$ Hence $\alpha=-3+\frac{6}{\sqrt{13}}, \qquad \beta=-3+\frac{9}{\sqrt{13}}.$

4. Compute $(\beta-\alpha)^2$

$\beta-\alpha=\left(-3+\frac{9}{\sqrt{13}}\right)-\left(-3+\frac{6}{\sqrt{13}}\right)=\frac{3}{\sqrt{13}}.$

Therefore, $(\beta-\alpha)^2=\left(\frac{3}{\sqrt{13}}\right)^2=\frac{9}{13}.$

So, $m=9,\quad n=13 \implies m+n=22.$

5. Compare with stored answer

Derived answer is 22, which matches option A.