The circle $4 x^2+4 y^2-12 x-12 y+9=0$

• Correct answer
  A

  touches both the axes
• B

  touches the $x$-axis only
• C

  touches the $y$-axis only
• D

  does not touch the axes

Solution

Given circle is $$ \begin{aligned} & & 4 x^2+4 y^2-12 x-12 y+9 & =0 \\ \Rightarrow & & x^2+y^2-3 x-3 y+\frac{9}{4} & =0 \\ \Rightarrow & & \left(x^2-3 x\right)+\left(y^2-3 y\right) & =-\frac{9}{4} \end{aligned} $$
$$ \begin{aligned} & \Rightarrow \quad\left(x-\frac{3}{2}\right)^2+\left(y-\frac{3}{2}\right)^2=\frac{9}{4} \\ & \Rightarrow \quad\left(x-\frac{3}{2}\right)^2+\left(y-\frac{3}{2}\right)^2=\left(\frac{3}{2}\right)^2 \end{aligned} $$ Hence, centre $=\left(\frac{3}{2}, \frac{3}{2}\right)$ and radius $=\frac{3}{2}$ So, the given circle touches both the axes.

Hence, the correct answer is option: 'A'.