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 polarised.
To form circularly polarised 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
$ |\vec{E}|^2=E_0+E_0+2 E_0 \cdot E_0 \cos \frac{\pi}{2}$
$ |\vec{E}|=E_0 \sqrt{2}=\text { constant }$
Hence, the correct graph will be (a).