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  4. The circle x2+y2 - 8x + 4y + 4 = 0 touches

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

KCETKCET 2004Conic Sections

Solution:

$
\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