Show that the circles ${\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{x}-2\mathrm{y}+1=0$; ${\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{x}-8\mathrm{y}+13=0$ touch each other. Find the points of contact and the equation of common tangent at the point of contact.

$$\begin{gathered} \mathrm{E}={\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{x}-2\mathrm{y}+1=0 {\mathrm{E}}^1={\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{x}-8\mathrm{y}+13=0 \\ {\mathrm{c}}_1=(3,1) {\mathrm{C}}_2=(-1,4) \\ {\mathrm{r}}_1=\sqrt{9+1-1}= 3 {\mathrm{r}}_2=\sqrt{1+16-13} \\ \overline{{\mathrm{C}}_1.{\mathrm{C}}_2}= \sqrt{{(-1-3)}^2{+(4-1)}^{2 }} {\mathrm{r}}_2=2 \\ \overline{{\mathrm{C}}_1.{\mathrm{C}}_2}= \sqrt{16+9} \\ \overline{{\mathrm{C}}_1}{\mathrm{C}}_2 =5 \end{gathered}$$


$$\begin{gathered} {\mathrm{r}}_1+{\mathrm{r}}_2=3+2\Rightarrow 5 \\ \overline{{\mathrm{C}}_1}{\mathrm{C}}_2 = {\mathrm{r}}_1+{\mathrm{r}}_2 \end{gathered}$$
$\therefore$Two circles are touch externally, then the point of contact is internal similitude then 
$$\begin{gathered} \mathrm{P}(\mathrm{x},\mathrm{y}) = \left(\frac{{\mathrm{mx}}_2+{\mathrm{nx}}_1}{\mathrm{m}+\mathrm{n}}, \frac{{\mathrm{my}}_2+{\mathrm{ny}}_1}{\mathrm{m}+\mathrm{n}}\right) \\ =\left(\frac{3(-1)+2\left(3\right)}{3+2}\text{'}\frac{3\left(4\right)+2\left(1\right)}{3+2}\right) \\ =\left(\frac{-3+6}{5},\frac{12+2}{5}\right) \\ =\left(\frac{3}{5},\frac{14}{5}\right) \end{gathered}$$
and equation of common target is 
$$\begin{gathered} \mathrm{s}-{\mathrm{s}}^1=0 \\ ({\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{x}-2\mathrm{y}+1)-({\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{x}-8\mathrm{y}+13)+0 \\ -8\mathrm{x}+6\mathrm{y}-12=0 \\ 4\mathrm{x}-3\mathrm{y}+6=0 \end{gathered}$$