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

x-axis

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y-axis

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Both axis

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Neither x-axis nor y-axis

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Solution

The correct option is D y-axis
$x^{2} + y^{2} - 8 x + 4 y + 4 = 0$
$\Rightarrow (x^{2} - 8 x + 16) + (y^{2} + 4 y + 4) = 16$
$\Rightarrow (x - 4)^{2} + (y + 2)^{2} = 4^{2}$
Center is $(4, - 2)$ and radius is $4$
Hence given circle touches y-axis

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