Two vertices of an equilateral triangle are (-1,0) and (1,0) and its third vertex lies above the X-axis. The equation of the circum circle of the triangle is

It is given that △ABC is an equilateral triangle, where the co-ordinates are A(−1,0) and B(1,0) respectively and the third vertex , say C, lies above the X-axis, C will lie on the Y-axis, as Y-axis is the perpendicular bisector of the line segment AB.

Let the co-ordinates of C be (0,k).
Length of side AB=2 units.

∴ Length of AC should also be 2 units

$$\begin{gathered} \Rightarrow AB=AC \\ \Rightarrow \sqrt{4}=\sqrt{1+k^2} \\ \Rightarrow k=\pm \sqrt{3} \\ \Rightarrow k=\sqrt{3}(\because k>0) \end{gathered}$$

3rd vertex, C=(0, $\sqrt{3}$)
For an equilateral triangle circumcentre = centroid = $\left(0,\frac{1}{\sqrt{3}}\right)$

and radius of circle=$\sqrt{{\left(1\right)}^2+{\left(\frac{1}{\sqrt{3}}\right)}^2}=\frac{2}{\sqrt{3}}$
circumcircle is 

$$\begin{gathered} (\mathrm{x}-0{)}^2+{\left(\mathrm{y}-\frac{1}{\sqrt{3}}\right)}^2={\left(\frac{2}{\sqrt{3}}\right)}^2 \\ \Rightarrow \sqrt{3}(x^2+y^2)-2y-\sqrt{3}=0 \end{gathered}$$