An equilateral triangle whose two vertices ar | vedclass

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(D) Let the vertices of the equilateral triangle be $A(-2, 0)$,$B(2, 0)$,and $C(0, 2\sqrt{3})$.The circumcenter of an equilateral triangle is the same

Vedclass · Accepted Answer

$\sqrt{3}x^2 + \sqrt{3}y^2 - 4y - 4\sqrt{3} = 0$

Vedclass · Answer

(D) Let the vertices of the equilateral triangle be $A(-2, 0)$,$B(2, 0)$,and $C(0, 2\sqrt{3})$.The circumcenter of an equilateral triangle is the same as its centroid.The centroid $G$ is given by $\left(\frac{-2+2+0}{3}, \frac{0+0+2\sqrt{3}}{3}\right) = \left(0, \frac{2\sqrt{3}}{3}\right) = \left(0, \frac{2}{\sqrt{3}}\right)$.The radius $R$ is the distance from the centroid to any vertex,e.g.,$B(2, 0)$:$R^2 = (2-0)^2 + (0 - \frac{2}{\sqrt{3}})^2 = 4 + \frac{4}{3} = \frac{16}{3}$.The equation of the circle is $(x-0)^2 + (y - \frac{2}{\sqrt{3}})^2 = \frac{16}{3}$.$x^2 + y^2 - \frac{4}{\sqrt{3}}y + \frac{4}{3} = \frac{16}{3}$.$x^2 + y^2 - \frac{4}{\sqrt{3}}y - 4 = 0$.Multiplying by $\sqrt{3}$,we get $\sqrt{3}x^2 + \sqrt{3}y^2 - 4y - 4\sqrt{3} = 0$.

An equilateral triangle whose two vertices are $(-2, 0)$ and $(2, 0)$ and which lies in the first and second quadrants only is circumscribed by a circle. Find the equation of this circle.

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