Q112 / 180 · COMEDK 2013 · Mathematics
The circle $x^{2}+y^{2}-8 x+4 y+4=0$ touches
- A$X$-axis
- Correct answerB$Y$-axis
- Cboth axes
- Dneither $X$-axis nor $Y$-axis.
Solution
We have, $x^{2}+y^{2}-8 x+4 y+4=0$
$\Rightarrow\left(x^{2}-8 x+16\right)+\left(y^{2}+4 y+4\right)=16$
$$
\Rightarrow \quad(x-4)^{2}+(y+2)^{2}=4^{2}
$$
$\therefore$ Centre is $(4,-2)$ and radius is 4 .
Hence, given circle touches $Y$-axis.
Hence, the correct answer is option: 'B'.