Show that the circles x²+y²– 4x – 6y – 12 = 0 and x² + y² + 6x + 18y + 26 = 0 touch each other. Find the point of contact and the equation of common tangent at the point of contact.

Given circles are
${\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-6\mathrm{y}-12=0$;
$\begin{array}{lc}{\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-6\mathrm{y}-12=0;& {\mathrm{x}}^2+{\mathrm{y}}^2+6\mathrm{x}+18\mathrm{y}+26=0\\ \mathrm{Centre} {\mathrm{C}}_1=(-\mathrm{g},-\mathrm{f})=(2,3)& \mathrm{Centre} {\mathrm{C}}_2=(-\mathrm{g},-\mathrm{f})=(-3,-9)\\ \mathrm{Radius} {\mathrm{r}}_1=\sqrt{{\mathrm{g}}^2+{\mathrm{f}}^2-\mathrm{c}}& \mathrm{Radius} {\mathrm{r}}_2=\sqrt{{\mathrm{g}}^2+{\mathrm{f}}^2-\mathrm{c}}\\ =\sqrt{4+9+12}=5& =\sqrt{9+81-26}=\sqrt{64}=8\end{array}$

Distance between the centre ${\mathrm{C}}_1{\mathrm{C}}_2=\sqrt{25+144}=\sqrt{169}=13$
$\begin{array}{l}{\mathrm{r}}_1+{\mathrm{r}}_2=5+8=13\\ {\mathrm{c}}_1{\mathrm{c}}_2={\mathrm{r}}_1+{\mathrm{r}}_2\end{array}$
$\therefore$The circles touch externally point of contact P divides c₁c₂ in the ratio r₁:r₂ internally in the ratio 5:8
$\begin{array}{l}\therefore \mathrm{p}=\frac{5{\mathrm{C}}_2+8{\mathrm{C}}_1}{5+8}=\frac{5(-3,-9)+8(2,3)}{13}=\frac{(-15,-45)+(16,24)}{13}\\ \left(\frac{-15+16}{13},\frac{-45+24}{13}\right)=\left(\frac{1}{13},\frac{-21}{13}\right)=\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)\end{array}$
Equation of common tangent at P is ${\mathrm{S}}_1=0\left(\text{ wrt }{1}^{\text{st }}\text{ circle or }{2}^{\text{nd }}\text{ circle }\right)$
$\begin{array}{l}\Rightarrow {\mathrm{xx}}_1+{\mathrm{yy}}_1-2\left(\mathrm{x}+{\mathrm{x}}_1\right)-3\left(\mathrm{y}+{\mathrm{y}}_1\right)-12=0\Rightarrow \mathrm{x}\left(\frac{1}{13}\right)+\mathrm{y}\left(\frac{-21}{13}\right)-2\left(\mathrm{x}+\frac{1}{13}\right)-3\left(\mathrm{y}-\frac{21}{13}\right)-12=0\\ \Rightarrow 5\mathrm{x}+12\mathrm{y}+19=0\end{array}$