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

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

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$. The center of the circle is $(-g, -f)$ and the radius is $\sqrt{g^2 + f^2 - c} = a$. Since the $x$-axis is a diameter,the center of the circle must lie on the $x$-axis. Therefore,the $y$-coordinate of the center must be zero,which implies $-f = 0$,so $f = 0$. Since the circle has the $x$-axis as a diameter,it passes through the origin if the center is on the $x$-axis and it touches the $x$-axis at the origin,but here it says $x$-axis is a diameter,meaning the circle intersects the $x$-axis at two points. For the $x$-axis to be a diameter,the center $(-g, 0)$ must lie on the $x$-axis (which is true for any $f=0$) and the distance from the center to the $x$-axis must be $0$. The radius is $a = \sqrt{g^2 + f^2 - c}$. Since $f = 0$,$a = \sqrt{g^2 - c}$. Squaring both sides,$a^2 = g^2 - c$,which implies $c = g^2 - a^2$. Looking at the options provided,none of the specific relations $f=2a, g=0, c=3a^2$ etc. are generally true for all such circles. Thus,the correct choice is $(D)$.

If the equation $x^2 + y^2 + 2gx + 2fy + c = 0$ represents a circle with the $x$-axis as a diameter and radius $a$,then

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