Question

The circle $x^{2}+y^{2} - 8x + 4y + 4 = 0$ touches

• A. x - axis
• B. y - axis
• C. both axis
• D. neither x-axis nor y-axis

Answer 1

Answer: (B)

$\begin{array}{l} x^{2}+y^{2}-8 x+4 y+4=0 \\ \Rightarrow\left(x^{2}-8 x+16\right)+\left(y^{2}+4 y+4\right)=16 \\ \Rightarrow(x-4)^{2}+(y+2)^{2}=4^{2} \end{array}$
Center is $(4,-2)$ and radius is 4
Hence given circle touches $y$-axis