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

x^{2} + y^{2} + 4 x = 0

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x^{2} + y^{2} - 4 y = 0

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x^{2} + y^{2} - 4 y = 12

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x^{2} + y^{2} + 4 x = 12

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Solution

The correct option is C $x^{2} + y^{2} - 4 y = 12$
Triangle $A B C$ is an equilateral triangle
Hence,
$B C = 4 \sqrt{3} 2 R = \frac{a}{sin A} 2 R = \frac{4 \sqrt{3}}{sin 60^{o}} = 8 R = 4$
Circumcentre of $A B C$ $(\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3}) = (0, 2)$
Equation of circumcircle
$(x - 0)^{2} + (y - 2)^{2} = 4^{2} x^{2} + y^{2} - 4 y = 12$

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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