Show that the circles ${\mathrm{x}}^2+{\mathrm{y}}^2=2\mathrm{x}=0$ and  ${\mathrm{x}}^2+{\mathrm{y}}^2+6\mathrm{x}-6\mathrm{y}+2=0$ touch each other. Find the coordinates of the point of contact. Is the point of contact external of internal ?

Given circles are
$\begin{array}{l}\mathrm{S}\equiv {\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}=0.\\ \Rightarrow \mathrm{g}=-1,\mathrm{f}=0,\mathrm{c}=0\end{array}$
Centre ${\mathrm{c}}_1=(-\mathrm{g},-\mathrm{f})=(1,0)$
             ${\mathrm{r}}_1=\sqrt{1+0-0}=1$
$\begin{array}{l}{\mathrm{S}}^{\mathrm{\prime }}\equiv {\mathrm{x}}^2+{\mathrm{y}}^2+6\mathrm{x}-6\mathrm{y}+2=0\\ \Rightarrow \mathrm{g}=3,{\mathrm{f}}^{\mathrm{\prime }}=-3,\mathrm{c}=2\end{array}$
Centre ${\mathrm{C}}_2=(-3,3), {\mathrm{r}}_2=\sqrt{9+9-2}=4.$
Now 
$\begin{array}{c}{c}_1{c}_2=\sqrt{(-3+1{)}^2+(3-0{)}^2}\\ =\sqrt{16+9}\\ =5\end{array}$
Now ${\mathrm{r}}_1+{\mathrm{r}}_2=1+4 = 5$
$\therefore {\mathrm{c}}_1{\mathrm{c}}_2={\mathrm{r}}_1+{\mathrm{r}}_2$
$\therefore$ Given circles touch each other externally.
Now point of contact P divides $\overline{{\mathrm{C}}_1{\mathrm{C}}_2}$ in the ratio ${\mathrm{r}}_1:{\mathrm{r}}_2$ internally then 
$\mathrm{P}=\left[\frac{{\mathrm{mx}}_2+{\mathrm{nx}}_1}{\mathrm{m}+\mathrm{n}},\frac{{\mathrm{my}}_2+{\mathrm{ny}}_1}{\mathrm{m}+\mathrm{n}}\right]$
$\begin{array}{l}\mathrm{P}=\left[\frac{1(-3)+4\left(1\right)}{5},\frac{1\left(3\right)+4\left(0\right)}{5}\right]\\ \mathrm{P}=\left[\frac{1}{5},\frac{3}{5}\right]\end{array}$