Explanation

(a) When two plane-polarised waves are superimposed, then under certain conditions, the resultant light vector rotates with a constant magnitude in a plane perpendicular to the direction of propagation. The tip of the vector traces a circle and the light is said to be circularly polarized.
To form circularly polarized light
$E_{x} = E_{0}\sin{\omega\ t}$
$E_{y} = E_{0}\cos{\omega\ t} = E_{0}\sin\left( \omega\ t + \frac{\pi}{2} \right)$
Where, $E_{0}$ is amplitude.
Resultant amplitude
$\left| \mathbf{E} \right|^{2} = E_{0}^{2} + E_{0}^{2} + 2E_{0}\ .E_{0}\cos\frac{\pi}{2}$
$\left| \mathbf{E} \right| = E_{0}\sqrt{2} = constant$
Hence, the correct graph will be option (a).