Let the shortest distance from (a, 0), a > 0, to the parabola $$y^{2}= 4x$$ be 4. Then the equation of the circle passing through the point (a,0) and the focus of the parabola, and having its centre on the axis of the parabola is:

The shortest distance from (a, 0) with a > 0 to the parabola $$y^2 = 4x$$ is specified as 4. A generic point on the parabola can be written as $$(t^2, 2t)$$, so the squared distance to $$(a,0)$$ is $$D^2 = (t^2 - a)^2 + 4t^2.$$ Differentiating and setting $$\frac{d(D^2)}{dt} = 0$$ gives $$t\bigl(4t^2 - 4a + 8\bigr) = 0.$$ Excluding the trivial solution $$t = 0$$ and requiring $$a > 2$$ yields $$t^2 = a - 2$$. Substituting back gives $$D^2 = 4 + 4(a - 2) = 4a - 4,$$ and imposing $$D = 4$$ leads to $$4a - 4 = 16$$, hence $$a = 5$$.

Having found $$a = 5$$, the required circle must pass through $$(5,0)$$ and the focus $$(1,0)$$ of the parabola, with its center on the x-axis at $$(h,0)$$. Equating distances from the center to these points, $$(5 - h)^2 = (1 - h)^2$$ leads to $$24 = 8h$$ and thus $$h = 3$$. The radius squared is then $$(5 - 3)^2 = 4$$, so the circle is $$(x - 3)^2 + y^2 = 4,$$ which expands to $$x^2 + y^2 - 6x + 5 = 0.$$

The correct answer is Option 2: $$x^2 + y^2 - 6x + 5 = 0$$.