If the circle x^2 + y^2 + 2gx + 2fy + c = 0 t | vedclass

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(B) The general equation of the circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.The center of the circle is $(-g, -f)$ and the radius $r$ is given by $r = \

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$g^2 = c$

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(B) The general equation of the circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.The center of the circle is $(-g, -f)$ and the radius $r$ is given by $r = \sqrt{g^2 + f^2 - c}$.If the circle touches the $x$-axis,the perpendicular distance from the center to the $x$-axis must be equal to the radius of the circle.The distance from the center $(-g, -f)$ to the $x$-axis $(y = 0)$ is $|-f| = |f|$.Therefore,$|f| = \sqrt{g^2 + f^2 - c}$.Squaring both sides,we get $f^2 = g^2 + f^2 - c$.Simplifying this,we get $g^2 = c$.

If the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ touches the $x$-axis,then:

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