Two vertices of an equilateral triangle are $(- 1, 0)$ and $(1, 0)$ and its third vertex lies above the $x -$ axis, an equation of the circumcircle is

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

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

$x^{2} + y^{2} - 2 y = 1$

none of these

answer is A.

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Detailed Solution

Let $A (- 1, 0), B (1, 0)$ and $C (0, b)$ be the vertices of the triangle, as C lies on the locus of points equidistance from $A (- 1, 0)$ and $B (1, 0)$ i.e., $y -$ axis.

Then $A B = A C \Rightarrow \sqrt{1 + b^{2}} = 2$
$\Rightarrow b^{2} = 3 \Rightarrow b = \sqrt{3} [∵ b > 0]$
Since the triangle is equilateral, the centre of the circumcircle is at the centroid of the triangle which is $(0, \sqrt{3} / 3) .$
Thus the equation of the circumcircle is
$\begin{array}{l} (x - 0)^{2} + (y - 1 / \sqrt{3})^{2} = (1 - 0)^{2} + (0 - 1 / \sqrt{3})^{2} \\ \Rightarrow x^{2} + y^{2} - (2 / \sqrt{3}) y + 1 / 3 = 1 + 1 / 3 \\ \Rightarrow 3 x^{2} + 3 y^{2} - 2 \sqrt{3} y = 3 . \end{array}$