Find the equations of the circles which touch 2x - 3 y + 1 = 0 at (1,1) and having radius $\sqrt{13}$.

$\text{ Normal form of }2\mathrm{x}-3\mathrm{y}+1=0\text{ is }$
$\Rightarrow 2\mathrm{x}-3\mathrm{y}=-1\Rightarrow -2\mathrm{x}+3\mathrm{y}=1$
$\text{ Divide by }\sqrt{4+9}=\sqrt{13}\because \text{ Given radius }\sqrt{13}$
$\text{ Point of contact }\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)=(1,1)$
$\begin{array}{l}\Rightarrow \frac{-2}{\sqrt{13}}\mathrm{x}+\frac{3}{\sqrt{13}}\mathrm{y}=\frac{1}{\sqrt{13}}\\ \Rightarrow \mathrm{xcos}\mathrm{\alpha }+\mathrm{ysin}\mathrm{\alpha }=\mathrm{P}\end{array}$
$\cos \mathrm{\alpha }=\frac{-2}{\sqrt{13}},\sin \mathrm{\alpha }=\frac{3}{\sqrt{13}},\mathrm{P}=\frac{1}{\sqrt{13}}$
$\text{ Center of the circles }=\left({\mathrm{x}}_1\pm \mathrm{rcos}\mathrm{\alpha },{\mathrm{y}}_1\pm \mathrm{rsin}\mathrm{\alpha }\right)$
$\begin{array}{l}=\left(1\pm \sqrt{16}\left(\frac{-2}{\sqrt{13}}\right),1+\sqrt{13}\left(\frac{3}{\sqrt{13}}\right)\right)\\ =(1-2,1+3)\mathrm{\&}(1+2,1-3)=(-1,4)\mathrm{\&}(3,-2)\end{array}$
$\text{ Required circles }\Rightarrow \mathrm{C}\left(\begin{array}{c}\underset{\mathrm{h}}{-1},\underset{\mathrm{k}}{4}\\ \end{array}\right)\mathrm{r}=\sqrt{13}$
$\begin{array}{l}\Rightarrow (\mathrm{x}-\mathrm{h}{)}^2+(\mathrm{y}-\mathrm{k}{)}^2={\mathrm{r}}^2\\ \Rightarrow (\mathrm{x}-1{)}^2+(\mathrm{y}-4{)}^2=(\sqrt{13}{)}^2\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{x}-8\mathrm{y}+4=0\\ \Rightarrow \mathrm{C}(3,-2),\mathrm{r}=\sqrt{13}\\ \Rightarrow (\mathrm{x}-3{)}^2+(\mathrm{y}+2{)}^2=(\sqrt{13}{)}^2\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{x}+4\mathrm{y}=0\end{array}$