Let the equation of the circle, which touches x-axis at the point (a,0),a > 0 and cuts off an intercept of length b on y-axis be $$x^{2}+y^{2}-\alpha x +\beta y+\gamma =0$$. If the circle lies below x-axis, then the ordered pair $$(2a,b^{2})$$ is equal to

A circle touching the x-axis at $$(a, 0)$$ has its centre at $$(a, r)$$ where $$r$$ is the radius. Since the circle lies below the x-axis, the centre is at $$(a, -r)$$ and the radius is $$r$$. Its equation is

$$(x - a)^2 + (y + r)^2 = r^2$$
$$x^2 + y^2 - 2ax + 2ry + a^2 = 0$$

Comparing this with the form $$x^2 + y^2 - \alpha x + \beta y + \gamma = 0$$, we identify $$\alpha = 2a$$, $$\beta = 2r$$, and $$\gamma = a^2$$. Thus, the first element of the ordered pair is $$2a = \alpha$$.

set $$x = 0$$ in the circle’s equation, yielding

$$y^2 + \beta y + \gamma = 0$$

The intercept of length $$b$$ on the y-axis satisfies

$$b = \sqrt{\beta^2 - 4\gamma}$$
$$b^2 = \beta^2 - 4\gamma$$

Hence, the ordered pair $$(2a, b^2)$$ becomes
$$ (2a, b^2) = (\alpha, \beta^2 - 4\gamma)$$

The correct answer is Option D: $$(\alpha, \beta^2 - 4\gamma)$$.