JEE Main 2025 — Circle Question with 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 $B(1,3)$
$\begin{aligned}
& \mathrm{AB}=\sqrt{16+36}=2 \sqrt{13} \\ & \mathrm{r}_1=3 \text { and } \mathrm{r}_2=2 \sqrt{13}-3 \\ & \mathrm{P}(\alpha, \beta), \alpha=\frac{\mathrm{r}_1(1)+\mathrm{r}_2(-3)}{\mathrm{r}_1+\mathrm{r}_2}, \beta=\frac{\mathrm{r}_1(3)+\mathrm{r}_2(-3)}{\mathrm{r}_1+\mathrm{r}_2} \\ & \alpha=\frac{3-3(2 \sqrt{13}-3)}{2 \sqrt{13}}, \beta=\frac{18-6 \sqrt{13}}{2 \sqrt{13}} \\ & (\beta-\alpha)^2=\left(\frac{6}{2 \sqrt{13}}\right)^2 \\ & (\beta-\alpha)^2=\left(\frac{6}{2 \sqrt{13}}\right)^2, \mathrm{~m}+\mathrm{n}=22
\end{aligned}$