Answer:

Explanation:
The energy of a photon is given by:

where
is the Planck constant
is the speed of light
is the wavelenght of the photon
For the microwave photons in this problem,

so their energy is

We are to show that the given parametric curve is a circle.
The trajectory of a circle with a radius r will satisfy the following relationship:

(with (x_c,y_c) being the center point)
We are given the x and y in a parametric form which can be further rewritten (using properties of sin/cos):

Squaring and adding both gives:

The last expression shows that the given parametric curve is a circle with the center (0,0) and radius A.