Show that the circles s =x² +y²–2x–4y –20= 0 and s' = x² + y² + 6x + 2y –90= 0 touch each other internally. Find the point of contact and the equation of the common tangent.

Given circles are
${\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-4\mathrm{y}-20=0$
${\text{ centre C}}_1=(-\mathrm{g},-\mathrm{f})=(1,2)$
$\text{ Radius }{\mathrm{r}}_1=\sqrt{{\mathrm{g}}^2+{\mathrm{f}}^2-\mathrm{c}}=\sqrt{1+4+20}=\sqrt{25}=5$
${\mathrm{x}}^2+{\mathrm{y}}^2+6\mathrm{x}+2\mathrm{y}-90=0$
${\text{ centre C}}_2=(-\mathrm{g},-\mathrm{f})=(-3,-1)$
$\text{ Radius }{\mathrm{r}}_2=\sqrt{{\mathrm{g}}^2+{\mathrm{f}}^2-\mathrm{c}}=\sqrt{9+1+90}=10$
$\text{ Distance between centers }{\mathrm{c}}_1{\mathrm{c}}_2=\sqrt{16+9}=5$
$\begin{array}{l}{\mathrm{r}}_1-{\mathrm{r}}_2=10-5=5\\ C_1{C}_2=\left|{\mathrm{r}}_1-{\mathrm{r}}_2\right|\end{array}$
$\therefore$The circles touch internally 
Point of contact p divides $\overline{C_1{C}_2}$ $\text{ in the ratio }{\mathrm{r}}_1:{\mathrm{r}}_2\text{ externally }$
$=-{\mathrm{r}}_1:{\mathrm{r}}_2=-5:10=-1:2$
$\begin{array}{l}\therefore \mathrm{P}=\frac{-1.C_2+2C_1}{-1+2}=\frac{-1(-3,-1)+2(1,2)}{1}\\ =(3+2,1+4)=(5,5)\end{array}$
Equation of common tangent at p(5, 5) to 1^st circle is s₁ = 0
$\begin{array}{l}\Rightarrow {\mathrm{xx}}_1+{\mathrm{yy}}_1-1\left(\mathrm{x}+{\mathrm{x}}_1\right)-2\left(\mathrm{y}+{\mathrm{y}}_1\right)-20=0\\ \Rightarrow \mathrm{x}\left(5\right)+\mathrm{y}\left(5\right)-1(\mathrm{x}+5)-2(\mathrm{y}+5)-20=0\\ \Rightarrow 5\mathrm{x}+5\mathrm{y}-\mathrm{x}-5-2\mathrm{y}-10-20=0\\ \Rightarrow 4\mathrm{x}+3\mathrm{y}-35=0\end{array}$