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(C) The lines are $y - \sqrt{3}x = 6$ and $y + \sqrt{3}x = 6$,and the $x$-axis is $y = 0$.For $y = 0$,the lines intersect the $x$-axis at $x = -2\sqrt

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$x^2 + y^2 - 4y = 12$

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(C) The lines are $y - \sqrt{3}x = 6$ and $y + \sqrt{3}x = 6$,and the $x$-axis is $y = 0$.For $y = 0$,the lines intersect the $x$-axis at $x = -2\sqrt{3}$ and $x = 2\sqrt{3}$.The intersection of the two lines $y - \sqrt{3}x = 6$ and $y + \sqrt{3}x = 6$ is $(0, 6)$.The vertices of the triangle are $A(0, 6)$,$B(-2\sqrt{3}, 0)$,and $C(2\sqrt{3}, 0)$.This is an isosceles triangle with base $BC$ on the $x$-axis.The circumcenter $(0, k)$ must be equidistant from $(0, 6)$ and $(2\sqrt{3}, 0)$.So,$|6 - k| = \sqrt{(2\sqrt{3} - 0)^2 + (0 - k)^2}$.$(6 - k)^2 = 12 + k^2$.$36 - 12k + k^2 = 12 + k^2$.$12k = 24$,so $k = 2$.The circumcenter is $(0, 2)$ and the radius $R = |6 - 2| = 4$.The equation of the circle is $(x - 0)^2 + (y - 2)^2 = 4^2$.$x^2 + y^2 - 4y + 4 = 16$.$x^2 + y^2 - 4y = 12$.

Find the equation of the circumcircle of the triangle formed by the lines $y \pm \sqrt{3}x = 6$ and the $x$-axis.

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