The circle x^2 + y^2 + 4x - 4y + 4 = 0 touche | vedclass

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(C) The given equation of the circle is $x^2 + y^2 + 4x - 4y + 4 = 0$. Comparing this with the general equation $x^2 + y^2 + 2gx + 2fy + c = 0$,we get

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$x$-axis and $y$-axis

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(C) The given equation of the circle is $x^2 + y^2 + 4x - 4y + 4 = 0$. Comparing this with the general equation $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $2g = 4 \implies g = 2$,$2f = -4 \implies f = -2$,and $c = 4$. The centre of the circle is $(-g, -f) = (-2, 2)$. The radius of the circle is $r = \sqrt{g^2 + f^2 - c} = \sqrt{2^2 + (-2)^2 - 4} = \sqrt{4 + 4 - 4} = \sqrt{4} = 2$. Since the absolute value of the $x$-coordinate of the centre is $|-2| = 2 = r$ and the absolute value of the $y$-coordinate of the centre is $|2| = 2 = r$,the circle touches both the $x$-axis and the $y$-axis.

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

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