Find the equation of the circle which touches ${\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}+6\mathrm{y}-12=0\text{ at }(-1,1)$ internally with a radius of 2

Given circle is                                             
${\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}+6\mathrm{y}-12=0$
$\text{ Centre }{\mathrm{C}}_1=(-\mathrm{g},-\mathrm{f})=(2,-3)$
$\text{ Radius }{\mathrm{r}}_1=\sqrt{{\mathrm{g}}^2+{\mathrm{f}}^2-\mathrm{c}}$
$=\sqrt{4+9+12}=5$
For the required circle
$\text{ Centre }{\mathrm{C}}_2=(\mathrm{h},\mathrm{k})$
$\text{ Radius }{\mathrm{r}}_2=2$

Required circle touches given circle internally at P(–1,1)
$\therefore$$\mathrm{P}\text{ divides }{\mathrm{C}}_1{\mathrm{C}}_2\text{ in the ratio }$
${\mathrm{C}}_1:{\mathrm{C}}_2\text{ externally }$
$\begin{array}{l}=-{\mathrm{r}}_1:{\mathrm{r}}_2=-5:2\Rightarrow \mathrm{P}=\frac{-5\cdot {\mathrm{C}}_2+2\cdot {\mathrm{C}}_1}{-5+2}\\ \Rightarrow (-1,1)=\frac{-5(\mathrm{h},\mathrm{k})+2(2,-3)}{-3}\\ =\left(\frac{-5\mathrm{h}+4}{-3},\frac{-5\mathrm{k}-6}{-3}\right)\Rightarrow \frac{-5\mathrm{h}+4}{-3}=-1\\ \Rightarrow \frac{-5\mathrm{k}-6}{-3}=1\Rightarrow \mathrm{h}=\frac{1}{5}\Rightarrow \mathrm{k}=\frac{-3}{5}\\ \therefore \text{ Centre }{\mathrm{C}}_2=\left(\frac{1}{5},\frac{-3}{5}\right)\text{ Radius }{\mathrm{r}}_2=2\\ \therefore \text{ Equation of the required circle is }\\ (\mathrm{x}-\mathrm{h}{)}^2+(\mathrm{y}-\mathrm{k}{)}^2={\mathrm{r}}^2\Rightarrow {\left(\mathrm{x}-\frac{1}{5}\right)}^2+{\left(\mathrm{y}+\frac{3}{5}\right)}^2=2^2\\ \Rightarrow 5{\mathrm{x}}^2+5{\mathrm{y}}^2-2\mathrm{x}+6\mathrm{y}-18=0\end{array}$