
Equation of circle x2+(y−(6−r))2=r2 touches 3x−y=0 $\begin{aligned}
& \mathrm{p}=\mathrm{r} \
& \frac{|0-(6-\mathrm{r})|}{2}=\mathrm{r} \
& |\mathrm{r}-6|=2 \mathrm{r} \
& \mathrm{r}=2
\end{aligned}\thereforeCircle\mathrm{x}^2+(\mathrm{y}-4)^2=4(2,4)$ Satisfies this equation