The circle $4x^2+4y^2-12x-12y+9=0$

$$\begin{gathered} Given circle 4{\mathrm{x}}^2+4{\mathrm{y}}^2-12\mathrm{x}-12\mathrm{y}+9=0 \\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-3\mathrm{x}-3\mathrm{y}+\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$4$}\right.=0 \\ Centre \left(C\right)=\left(\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right., \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right), \mathrm{r}=\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ Here r=\left|x-coordinate of centre\right|=\left|y-cordinate of centre\right| \\ So,circle touches both the axes \end{gathered}$$