A circle touches the $$x$$-axis and also touches the circle with centre at $$(0, 3)$$ and radius $$2$$. The locus of the centre of the circle is

Let the centre of the required circle be $$C(h,k)$$ and its radius be $$r$$.

Because the circle touches the $$x$$-axis, the distance of its centre from the $$x$$-axis equals its radius:
$$k = r \qquad -(1)$$

The given circle has centre $$O(0,3)$$ and radius $$2$$. For the two circles to touch each other, the distance $$OC$$ must equal the sum or the difference of their radii.

The distance between the centres is
$$OC = \sqrt{h^2 + (k-3)^2} \qquad -(2)$$

If the circles touch externally, then $$OC = r + 2$$ Using $$r=k$$ from (1): $$\sqrt{h^2 + (k-3)^2} = k + 2$$

Squaring both sides, $$h^2 + (k-3)^2 = (k+2)^2$$ $$h^2 + k^2 - 6k + 9 = k^2 + 4k + 4$$ Cancelling $$k^2$$ and rearranging, $$h^2 = 10k - 5 \qquad -(3)$$

Equation (3) represents a parabola whose axis is the $$k$$-axis, vertex at $$(0,\tfrac{1}{2})$$, and opening upwards.

For internal contact we would need $$OC = |r - 2| = |k - 2|.$$ If the smaller circle were inside the bigger one, we would have $$k \lt 2$$, giving $$\sqrt{h^2 + (k-3)^2} = 2 - k.$$ Squaring, $$h^2 + (k-3)^2 = (2 - k)^2 \implies h^2 = 2k - 5.$$ But for $$k \lt 2$$, the right side $$2k-5$$ is negative, which is impossible for $$h^2$$. Hence internal tangency is not feasible.

Therefore the only admissible locus is given by (3): $$h^2 = 10k - 5.$$

Since this equation is of the form $$x^2 = 4a(y - y_0)$$, it represents a parabola.

Hence, the locus of the centre is a parabola.

Option D which is: a parabola