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

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

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touches both axes

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(B) The given equation of the circle is $x^2 + y^2 - 4x - 4y + 4 = 0$.Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -2$,$f = -2$,and $c = 4$.The center 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 distance from the center $(2, 2)$ to both the $x$-axis and $y$-axis is equal to the radius $r = 2$,the circle touches both axes.

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

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