The equation of the circle of radius 3 that lies in $4^{th}$ quadrant and touching the lines x = 0, y = 0 is 
 

$$\begin{gathered} \mathrm{Since} \mathrm{given} \mathrm{circle} \mathrm{with} \mathrm{radius} 3 \mathrm{lies} \mathrm{in} \mathrm{fourth} \mathrm{quadrant} \mathrm{and} \mathrm{touches} \mathrm{both} \mathrm{the} \mathrm{axes} \\ \mathrm{then} \mathrm{centre}\left(\mathrm{C}\right)=(3, -3), \mathrm{r}=3 \\ Equation of required circle is {(\mathrm{x}-3)}^2+{(\mathrm{y}+3)}^2=9 \\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{x}+6\mathrm{y}+9=0 \end{gathered}$$