The circle(s) touching the x-axis at a distan | vedclass

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(D) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.Since the circle touches the $x$-axis at $(3, 0)$ or $(-3, 0)$,the center is $(\

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$(A, C)$

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(D) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.Since the circle touches the $x$-axis at $(3, 0)$ or $(-3, 0)$,the center is $(\pm 3, -f)$ and the radius is $|f|$.Given the circle touches the $x$-axis at a distance of $3$ from the origin,the center is $(\pm 3, f)$. The condition for touching the $x$-axis is $g^2 = c$.Since it touches at $x = 3$ or $x = -3$,the $x$-coordinate of the center is $3$ or $-3$,so $-g = 3$ or $-g = -3$,implying $g = -3$ or $g = 3$. In both cases,$g^2 = 9$,so $c = 9$.The intercept on the $y$-axis is given by $2 \sqrt{f^2 - c} = 2 \sqrt{7}$,which implies $f^2 - c = 7$.Substituting $c = 9$,we get $f^2 - 9 = 7$,so $f^2 = 16$,which means $f = \pm 4$.The equation of the circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.Substituting $g = -3$ (or $3$) and $f = \pm 4$ and $c = 9$,we get $x^2 + y^2 - 6x \pm 8y + 9 = 0$.Thus,the circles are $x^2 + y^2 - 6x + 8y + 9 = 0$ and $x^2 + y^2 - 6x - 8y + 9 = 0$.

The circle$(s)$ touching the $x$-axis at a distance of $3$ from the origin and having an intercept of length $2 \sqrt{7}$ on the $y$-axis is (are):

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