Find the equation of the circle which touches the circle x² + y² – 2x – 4y – 20 = 0 externally at (5, 5) with radius 5 units

Given circle is ${\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-4\mathrm{y}-20=0$
$\text{ Centre }{\mathrm{c}}_1=(1,2)\text{ for the required circle, we }$
$\text{ Radius }{\mathrm{r}}_1=\sqrt{1+4+20}\text{ center }{\mathrm{c}}_2=(\mathrm{h},\mathrm{k})=5$
$\text{ Radius }{\mathrm{r}}_2=5\text{ units }$
Required circle touches given circle externally at p(5, 5)
$\mathrm{P}\text{ divides }{\mathrm{c}}_1{\mathrm{c}}_2\text{ in the }{\mathrm{r}}_1:{\mathrm{r}}_2\text{ internally }=5:5=1:1$
$\mathrm{P}=\text{ mid point of }{\mathrm{c}}_1{\mathrm{c}}_2$
$=\left(\frac{1+\mathrm{h}}{2},\frac{2+\mathrm{k}}{2}\right)\Rightarrow (5,5)=\left(\frac{1+\mathrm{h}}{2},\frac{2+\mathrm{k}}{2}\right)$

$$\begin{gathered} \Rightarrow \frac{1+\mathrm{h}}{2} = 5 \Rightarrow \frac{2+\mathrm{k}}{2} = 5 \\ \Rightarrow \mathrm{h}+1=10 \Rightarrow \mathrm{k}+2=10 \\ \Rightarrow \mathrm{h} = 9 \Rightarrow \mathrm{k} = 8 \end{gathered}$$
$\mathrm{Centre} {\mathrm{c}}_2=(9,8) \mathrm{Radius} {\mathrm{r}}_2 = 5$
$\therefore$Equation of required circle is 
$\begin{array}{l}(\mathrm{x}-\mathrm{h}{)}^2+(\mathrm{y}-\mathrm{k}{)}^2={\mathrm{r}}_2^2\Rightarrow (\mathrm{x}-9{)}^2+(\mathrm{y}-8{)}^2=25\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-18\mathrm{x}-16\mathrm{y}+81+64=25\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-18\mathrm{x}-16\mathrm{y}+120=0\end{array}$