The circle $4 x^2+4 y^2-12 x-12 y+9=0$
- touches both the axes
- touches the $x$-axis only
- touches the $y$-axis only
- does not touch the axes
Solution

$ \begin{aligned} & \Rightarrow \quad\left(x-\frac{3}{2}\right)^2+\left(y-\frac{3}{2}\right)^2=\frac{9}{4} \\ & \Rightarrow \quad\left(x-\frac{3}{2}\right)^2+\left(y-\frac{3}{2}\right)^2=\left(\frac{3}{2}\right)^2 \end{aligned} $ Hence, centre $=\left(\frac{3}{2}, \frac{3}{2}\right)$ and radius $=\frac{3}{2}$ So, the given circle touches both the axes
Asked in: AP EAMCET 2013