Find the equation of a circle whose center li | vedclass

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(A) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.Since the circle passes through the origin $(0, 0)$,we have $c = 0$.The center o

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$x^2 + y^2 + 6y = 0$

Vedclass · Answer

(A) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.Since the circle passes through the origin $(0, 0)$,we have $c = 0$.The center of the circle is $(-g, -f)$.Since the center lies on the $y$-axis,the $x$-coordinate of the center must be zero,so $-g = 0$,which implies $g = 0$.The equation of the circle becomes $x^2 + y^2 + 2fy = 0$.The radius of the circle is given by $\sqrt{g^2 + f^2 - c} = \sqrt{0^2 + f^2 - 0} = |f|$.Given that the radius is $3$,we have $|f| = 3$,which means $f = 3$ or $f = -3$.Substituting $f = 3$,we get $x^2 + y^2 + 6y = 0$.Substituting $f = -3$,we get $x^2 + y^2 - 6y = 0$.Comparing with the given options,$x^2 + y^2 + 6y = 0$ is the correct choice.

Find the equation of a circle whose center lies on the $y$-axis,has a radius of $3$,and passes through the origin.

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