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Question

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 


A
x2+y2+4x=0
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B
x2+y24y=0
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C
x2+y24y=12
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D
x2+y2+4x=12
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Solution

The correct option is C $$x^2+y^2-4y=12$$
Triangle $$ABC$$ is an equilateral triangle
Hence,
$$BC=4\sqrt{3} \\ 2R=\cfrac{a}{\sin A} \\ 2R=\cfrac{4\sqrt{3}}{\sin 60^o}=8 \\ R=4$$
Circumcentre of $$ABC$$ $$\left( \cfrac{x_1+x_2+x_3}{3},\cfrac{y_1+y_2+y_3}{3} \right) \\ =(0,2) $$
Equation of circumcircle
$$(x-0)^2+(y-2)^2=4^2 \\ x^2+y^2-4y=12$$

901754_698705_ans_f11c3ae82975447d876ea962c0f135a7.png

Mathematics

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