Write the general plane wave as the sum of two orthogonal components and every polarisation state falls out of it:

\(\vec{E}=\hat{a}_{x}E_{x}\cos(\omega t-\beta z)+\hat{a}_{y}E_{y}\cos(\omega t-\beta z+\delta)\)

Only two quantities matter — the amplitude ratio \(E_{y}/E_{x}\) and the phase difference \(\delta\). The tip of the resultant vector traces a figure in the transverse plane, and that figure is in general an ellipse:

So circular polarisation is indeed the special case in which the ellipse degenerates to a circle, requiring equal amplitudes and a quarter-cycle phase difference. Linear polarisation is the opposite degenerate case, the ellipse collapsed to a line. Elliptical is the parent state and the other two are its limits — (A) is true.

The reason is true as well and does account for the assertion. Two orthogonal components with a phase difference are exactly what makes the electric vector rotate rather than merely oscillate along a fixed direction; the rotating vector is what draws the ellipse, and the circle is one particular ellipse it can draw. Because the reason supplies the mechanism from which the assertion follows, the code is 1. The reason is somewhat loosely worded — it does not spell out that a phase difference is essential, since without one the "rotation" collapses to a line — but the intended sense is clear.

The measure of how elliptical a wave is is the axial ratio

\(AR=\dfrac{\text{major axis}}{\text{minor axis}}\qquad 1\le AR\le\infty\)

with AR = 1 for perfect circular polarisation and infinite for linear.

Why circular polarisation is used. A circularly polarised link is insensitive to the relative orientation of transmitter and receiver, which is why it is standard in satellite communication and GPS, where a spinning or tumbling spacecraft would otherwise cause deep polarisation-mismatch fades. It also survives Faraday rotation in the ionosphere, which would rotate a linearly polarised wave out of alignment.

Hence, both (A) and (R) are true and (R) is the correct explanation of (A).