Show that the circles ${\mathrm{x}}^2+{\mathrm{y}}^2=6\mathrm{x}-9\mathrm{y}+13=0,{\mathrm{x}}^2+{\mathrm{y}}^2=2\mathrm{x}-16\mathrm{y}=0$touch each other. Also find
the point of contact and common tangent at this point of contact
 

Given circle are ${\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{x}-9\mathrm{y}+13=0;{\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-16\mathrm{y}=0$$\begin{array}{cc}\text{ Centre }{\mathrm{C}}_1=(-\mathrm{g},-\mathrm{f})& \text{ Centre }{\mathrm{C}}_2=(-\mathrm{g},-\mathrm{f})\\ =\left(3,\frac{9}{2}\right)& =(1,8)\\ \text{ Radius }{\mathrm{r}}_1=\sqrt{{\mathrm{g}}^2+{\mathrm{f}}^2-\mathrm{c}}& \text{ Radius }{\mathrm{r}}_2=\sqrt{{\mathrm{g}}^2+{\mathrm{f}}^2-\mathrm{c}}\\ \begin{array}{c}{r}_1=\sqrt{9+\frac{81}{4}-13}\\ =\frac{\sqrt{65}}{2}\end{array}& \begin{array}{l}{\mathrm{r}}_2=\sqrt{{\mathrm{g}}^2+{\mathrm{f}}^2-\mathrm{c}}\\ =\sqrt{1+64}=\sqrt{65}\end{array}\end{array}$
Distance between the centers${\mathrm{C}}_1{\mathrm{C}}_2=\sqrt{4+\frac{49}{4}}=\frac{\sqrt{65}}{2}$
${\mathrm{r}}_2-{\mathrm{r}}_1=\sqrt{65}-\frac{\sqrt{65}}{2}=\frac{\sqrt{65}}{2}{\mathrm{C}}_1{\mathrm{C}}_2=\left|{\mathrm{r}}_2-{\mathrm{r}}_1\right|$
$=-{\mathrm{r}}_1:{\mathrm{r}}_2=-\sqrt{\frac{65}{2}}:\sqrt{65}=-\frac{1}{2}:1=-1:2\therefore \mathrm{P}=\frac{-1{\mathrm{C}}_2+2{\mathrm{C}}_1}{-1+2}$

$\therefore$The circles touch internally then  point of contact P divides C₁C₂ in the ratio r₁:r₂ externally
$=\frac{-1(1,8)+2\left(3,\frac{9}{2}\right)}{1}\Rightarrow \frac{(-1,-8)+(6,9)}{1}=(5,1)$
Equation of common tangent at P (5,1) w.r.t. 1st circle is S₁= 0
$\begin{array}{l}\Rightarrow {\mathrm{xx}}_1+{\mathrm{yy}}_1-3\left(\mathrm{x}+{\mathrm{x}}_1\right)-\frac{9}{2}\left(\mathrm{y}+{\mathrm{y}}_1\right)+13=0\Rightarrow \mathrm{x}\left(5\right)+\mathrm{y}\left(1\right)-3(\mathrm{x}+5)-\frac{9}{2}(\mathrm{y}+1)+13=0\\ \Rightarrow 4\mathrm{x}-7\mathrm{y}-13=0\end{array}$