Equation of a circle touching the line $|\mathrm{x}-1|+|\mathrm{y}-4|=6$ is 

$$\begin{gathered} \left|\mathrm{x}-1\right|+\left|\mathrm{y}-4\right|=6 \\ \Rightarrow \left|\mathrm{X}\right|+\left|\mathrm{Y}\right|=6, \mathrm{wherever} \mathrm{x}-1=\mathrm{X}, \mathrm{y}-4=\mathrm{Y} \\ \Rightarrow \pm X\pm Y=6 \mathrm{are} \mathrm{tangents} \mathrm{to} \mathrm{a} \mathrm{circle} \mathrm{whose} \mathrm{centre} =(0,0) \\ \mathrm{r}=\mathrm{d}=\frac{\left|6\right|}{\sqrt{1^2+1^2}}=\frac{6}{\sqrt{2}}=3\sqrt{2} \end{gathered}$$
Equation of circle ${\mathrm{X}}^2+Y^2=18\Rightarrow {\left(\mathrm{x}-1\right)}^2+{\left(\mathrm{y}-4\right)}^2=18$