Equation of a circle is  $\mathrm{S}(\mathrm{x},\mathrm{y})=0,$ such that $\mathrm{S}(2,3)=16$ If the circle  $\mathrm{S}(\mathrm{x},\mathrm{y})=0,$touches the line $3\mathrm{x}+4\mathrm{y}-7=0$ at (1, 1) then the equation of the circle  is given by

Any circle which touches $3\mathrm{x}+4\mathrm{y}-7=0\text{ at }(1,1)$ will be of the form
$\mathrm{S}(\mathrm{x},\mathrm{y})\equiv (\mathrm{x}-1{)}^2+(\mathrm{y}-1{)}^2+\mathrm{\lambda }(3\mathrm{x}+4\mathrm{y}-7)=0$
Since    $\mathrm{S}(2,3)=16\Rightarrow 1+4+\mathrm{\lambda }\left(11\right)=16$
$\therefore \mathrm{\lambda }=1$
So, required circle will be ${\mathrm{x}}^2+{\mathrm{y}}^2+\mathrm{x}+2\mathrm{y}-5=0$