A circle cuts a chord of length 4a on the x-axis and passes through a point on the y-axis, distant 2b from the origin. Then the locus of the centre of this circle, is:

Let us denote the centre of the required circle by the coordinates $$(h,k)$$ and its radius by $$r$$.

Hence the general equation of the circle is

$$ (x-h)^2 + (y-k)^2 = r^2. $$

We translate both given geometric conditions into algebraic equations.

First condition: the circle cuts a chord of length $$4a$$ on the $$x$$-axis. On the $$x$$-axis we have $$y=0$$, so substituting $$y=0$$ in the equation of the circle gives

$$ (x-h)^2 + k^2 = r^2. $$

This is a quadratic in $$x$$ whose roots give the $$x$$-coordinates of the points of intersection with the $$x$$-axis. Solving for $$x$$ gives

$$ x = h \pm \sqrt{\,r^2 - k^2\,}. $$

The distance between these two intersection points is therefore

$$ |(h+\sqrt{r^2-k^2})-(h-\sqrt{r^2-k^2})| = 2\sqrt{\,r^2-k^2\,}. $$

But we are told that this distance equals the given chord length $$4a$$, so

$$ 2\sqrt{\,r^2-k^2\,} = 4a \quad\Longrightarrow\quad \sqrt{\,r^2-k^2\,} = 2a \quad\Longrightarrow\quad r^2 - k^2 = 4a^2. \quad -(1) $$

Second condition: the circle passes through the point on the $$y$$-axis that is at a distance $$2b$$ from the origin. That point has coordinates $$(0,\,2b)$$. Substituting $$(x,y)=(0,2b)$$ in the circle’s equation gives

$$ (0-h)^2 + (2b-k)^2 = r^2 \quad\Longrightarrow\quad h^2 + (2b-k)^2 = r^2. \quad -(2) $$

Now we eliminate $$r^2$$ between equations (1) and (2). From (1) we have $$r^2 = k^2 + 4a^2$$. Putting this value into (2) yields

$$ h^2 + (2b-k)^2 = k^2 + 4a^2. $$

Expanding the square and simplifying step by step:

$$ h^2 + (2b-k)^2 = h^2 + (4b^2 - 4bk + k^2). $$

So the left-hand side becomes

$$ h^2 + 4b^2 - 4bk + k^2. $$

Equating this to the right-hand side $$k^2 + 4a^2$$ and cancelling the identical $$k^2$$ terms on both sides, we obtain

$$ h^2 + 4b^2 - 4bk = 4a^2. $$

Collecting all terms to one side we get the Cartesian relation connecting $$h$$ and $$k$$:

$$ h^2 - 4bk + 4b^2 - 4a^2 = 0. $$

To write the locus, we now replace the particular centre coordinates $$(h,k)$$ by the general point $$(x,y)$$:

$$ x^2 - 4by + 4b^2 - 4a^2 = 0. $$

Rearranging to the more familiar form,

$$ x^2 = 4by - 4(b^2 - a^2). $$

This equation contains the squared term $$x^2$$ but no $$y^2$$ term, which is the defining algebraic characteristic of a parabola (a conic with eccentricity one). Hence the centre of the circle moves along a parabolic curve.

Hence, the correct answer is Option D.