Let $C_1$ be the circle in the third quadrant of radius 3 , that touches both coordinate axes. Let $\mathrm{C}_2$ be the circle with centre $(1,3)$ that touches $\mathrm{C}_1$ externally at the point $(\alpha, \beta)$. If $(\beta-\alpha)^2=\frac{m}{n}, \operatorname{gcd}(m, n)=1$, then $m+n$ is equal to :

• A

  9
• B

  13
• Correct answer
  C

  22
• D

  31

Solution

$C_1:(x+3)^2+(y+3)^2=3^2$

Let $\mathrm{C}_1$ and $\mathrm{C}_2$ has centres $A\left(-3_1-3\right)$ and $B(1,3)$
$\begin{aligned}<br/>& \mathrm{AB}=\sqrt{16+36}=2 \sqrt{13} \\ & \mathrm{r}_1=3 \text { and } \mathrm{r}_2=2 \sqrt{13}-3 \\ & \mathrm{P}(\alpha, \beta), \alpha=\frac{\mathrm{r}_1(1)+\mathrm{r}_2(-3)}{\mathrm{r}_1+\mathrm{r}_2}, \beta=\frac{\mathrm{r}_1(3)+\mathrm{r}_2(-3)}{\mathrm{r}_1+\mathrm{r}_2} \\ & \alpha=\frac{3-3(2 \sqrt{13}-3)}{2 \sqrt{13}}, \beta=\frac{18-6 \sqrt{13}}{2 \sqrt{13}} \\ & (\beta-\alpha)^2=\left(\frac{6}{2 \sqrt{13}}\right)^2 \\ & (\beta-\alpha)^2=\left(\frac{6}{2 \sqrt{13}}\right)^2, \mathrm{~m}+\mathrm{n}=22<br/>\end{aligned}$

Hence, the correct answer is option: 'C'.