The equation of the circumcircle of the trian | vedclass

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(C) The given lines are $L_1: y + \sqrt{3}x = 6$,$L_2: y - \sqrt{3}x = 6$,and $L_3: y = 0$.Finding the vertices of the triangle:Intersection of $L_1$

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

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(C) The given lines are $L_1: y + \sqrt{3}x = 6$,$L_2: y - \sqrt{3}x = 6$,and $L_3: y = 0$.Finding the vertices of the triangle:Intersection of $L_1$ and $L_2$: $y = 6, x = 0$. Vertex $A = (0, 6)$.Intersection of $L_1$ and $L_3$: $y = 0, \sqrt{3}x = 6 \implies x = 2\sqrt{3}$. Vertex $B = (2\sqrt{3}, 0)$.Intersection of $L_2$ and $L_3$: $y = 0, -\sqrt{3}x = 6 \implies x = -2\sqrt{3}$. Vertex $C = (-2\sqrt{3}, 0)$.Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.Since the circle passes through $(0, 6)$,$(2\sqrt{3}, 0)$,and $(-2\sqrt{3}, 0)$:For $(2\sqrt{3}, 0)$: $12 + 4g\sqrt{3} + c = 0$For $(-2\sqrt{3}, 0)$: $12 - 4g\sqrt{3} + c = 0$Adding these gives $24 + 2c = 0 \implies c = -12$. Then $4g\sqrt{3} = 0 \implies g = 0$.For $(0, 6)$: $0^2 + 6^2 + 2f(6) - 12 = 0 \implies 36 + 12f - 12 = 0 \implies 12f = -24 \implies f = -2$.The equation is $x^2 + y^2 - 4y - 12 = 0$,which is $x^2 + y^2 - 4y = 12$.

The equation of the circumcircle of the triangle formed by the lines $y + \sqrt{3}x = 6$,$y - \sqrt{3}x = 6$,and $y = 0$ is:

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