An ellipse is drawn by taking a diameter of the circle $$(x-1)^2 + y^2 = 1$$ as its semiminor axis and a diameter of the circle $$x^2 + (y-2)^2 = 4$$ as its semi-major axis. If the centre of the ellipse is the origin and its axes are the coordinate axes, then the equation of the ellipse is

The given data are the lengths of two diameters that will serve as the semi-axes of the required ellipse.

Circle 1: $$(x-1)^2 + y^2 = 1$$ has radius $$1$$, so its diameter is $$2$$. This diameter is to be used as the semiminor axis, hence $$b = 2 \; \Longrightarrow \; b^{2} = 4.$$

Circle 2: $$x^{2} + (y-2)^2 = 4$$ has radius $$2$$, so its diameter is $$4$$. This diameter is to be used as the semimajor axis, hence $$a = 4 \; \Longrightarrow \; a^{2} = 16.$$

The centre of the ellipse is the origin and its axes coincide with the coordinate axes. A standard-position ellipse therefore has the equation

$$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1.$$

Substituting $$a^{2}=16,\; b^{2}=4$$ gives

$$\frac{x^{2}}{16} + \frac{y^{2}}{4} = 1 \; \Longrightarrow \; x^{2} + 4y^{2} = 16.$$

Hence the required ellipse is represented by Option D which is: $$x^{2} + 4y^{2} = 16$$.