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

Solution

## The correct option is C $$x^2+y^2-4y=12$$Triangle $$ABC$$ is an equilateral triangleHence,$$BC=4\sqrt{3} \\ 2R=\cfrac{a}{\sin A} \\ 2R=\cfrac{4\sqrt{3}}{\sin 60^o}=8 \\ R=4$$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$$(x-0)^2+(y-2)^2=4^2 \\ x^2+y^2-4y=12$$Mathematics

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