If two vertices of an equilateral triangle are $$A(-a, 0)$$ and $$B(a, 0)$$, $$a > 0$$, and the third vertex C lies above x-axis then the equation of the circumcircle of $$\triangle ABC$$ is :

We have the two given vertices of the equilateral triangle as $$A(-a,0)$$ and $$B(a,0)$$ with $$a>0$$. The side $$AB$$ therefore lies on the $$x$$-axis and its length is

$$AB=\sqrt{(a-(-a))^{2}+(0-0)^{2}}=\sqrt{(2a)^{2}}=2a.$$

Let the third vertex be $$C(x_c,y_c)$$ with $$y_c>0$$ so that it is above the $$x$$-axis. Because the triangle is equilateral, the distances $$AC$$ and $$BC$$ must also be $$2a$$. So we write

$$AC^{2}=(x_c+a)^{2}+y_c^{2}= (2a)^{2}=4a^{2},$$

$$BC^{2}=(x_c-a)^{2}+y_c^{2}= (2a)^{2}=4a^{2}.$$

Subtracting the second equation from the first eliminates $$y_c^{2}$$ and gives

$$(x_c+a)^{2}-(x_c-a)^{2}=0.$$

Expanding both squares, we obtain

$$\bigl(x_c^{2}+2ax_c+a^{2}\bigr)-\bigl(x_c^{2}-2ax_c+a^{2}\bigr)=0 \;\Longrightarrow\;4ax_c=0.$$

Since $$a\neq0$$, we must have $$x_c=0$$. Thus the third vertex lies on the $$y$$-axis, and we now find its $$y$$-coordinate from either length equation:

$$a^{2}+y_c^{2}=4a^{2}\;\Longrightarrow\;y_c^{2}=3a^{2}\;\Longrightarrow\;y_c=\sqrt3\,a.$$

Hence the coordinates of $$C$$ are $$C(0,\sqrt3\,a).$$

For an equilateral triangle, the circumcenter (center of the circumscribed circle) coincides with the intersection of the perpendicular bisectors and therefore lies on the symmetry axis $$x=0$$. Let that center be $$O(0,h)$$ and its radius be $$R$$. Because all vertices are equally distant from the circumcenter, we write the distance formula for vertex $$A$$:

$$OA^{2}=(-a-0)^{2}+(0-h)^{2}=a^{2}+h^{2}=R^{2}.$$

In an equilateral triangle of side $$s$$, the circumradius is known to be $$R=\dfrac{s}{\sqrt3}.$$ Here $$s=2a$$, so

$$R=\dfrac{2a}{\sqrt3}\quad\Longrightarrow\quad R^{2}=\dfrac{4a^{2}}{3}.$$

Substituting this $$R^{2}$$ into $$a^{2}+h^{2}=R^{2}$$, we get

$$a^{2}+h^{2}=\dfrac{4a^{2}}{3} \;\Longrightarrow\; h^{2}=\dfrac{4a^{2}}{3}-a^{2}=\dfrac{a^{2}}{3} \;\Longrightarrow\; h=\dfrac{a}{\sqrt3}.$$

Thus the circumcenter is $$O\!\left(0,\dfrac{a}{\sqrt3}\right)$$ and the radius remains $$R=\dfrac{2a}{\sqrt3}.$$

The general equation of a circle with centre $$(x_0,y_0)$$ and radius $$R$$ is

$$(x-x_0)^{2}+(y-y_0)^{2}=R^{2}.$$

Placing $$x_0=0$$, $$y_0=\dfrac{a}{\sqrt3}$$ and $$R^{2}=\dfrac{4a^{2}}{3}$$ we write

$$x^{2}+\left(y-\dfrac{a}{\sqrt3}\right)^{2}=\dfrac{4a^{2}}{3}.$$

Expanding the square in $$y$$, we have

$$x^{2}+y^{2}-2y\dfrac{a}{\sqrt3}+\dfrac{a^{2}}{3}=\dfrac{4a^{2}}{3}.$$

Now we collect like terms on the left:

$$x^{2}+y^{2}-\dfrac{2a}{\sqrt3}\,y+\dfrac{a^{2}}{3}-\dfrac{4a^{2}}{3}=0 \;\Longrightarrow\; x^{2}+y^{2}-\dfrac{2a}{\sqrt3}\,y-a^{2}=0.$$

Multiplying the entire equation by $$3$$ to clear the denominators gives

$$3x^{2}+3y^{2}-2\sqrt3\,a\,y-3a^{2}=0.$$

Rearranging with the constant term on the right side yields the standard form presented in the options:

$$3x^{2}+3y^{2}-2\sqrt3\,a\,y=3a^{2}.$$

This exactly matches Option A.

Hence, the correct answer is Option A.