If the radius of the circle x^2 + y^2 + 2gx + | vedclass

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(C) The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.Its center is $(-g, -f)$ and its radius is $r = \sqrt{g^2 + f^2 - c}$.For the

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$g^2 = f^2 = c = r^2$

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(C) The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.Its center is $(-g, -f)$ and its radius is $r = \sqrt{g^2 + f^2 - c}$.For the circle to touch both axes,the distance from the center to both axes must be equal to the radius.Thus,$|-g| = |-f| = r$,which implies $|g| = |f| = r$.Also,the radius formula gives $r^2 = g^2 + f^2 - c$.Substituting $g^2 = r^2$ and $f^2 = r^2$,we get $r^2 = r^2 + r^2 - c$,which simplifies to $c = r^2$.Therefore,the condition is $g^2 = f^2 = c = r^2$.

If the radius of the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is $r$,then it will touch both the axes if:

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