Imagine one of the point in the wave. It is in oscillation. So its displacement can be written as $Y = A\sin(\theta)$ where $\theta= \omega t$. $$Y(t)=A\sin(\omega t) \tag{1}$$
Time for one wave length $\lambda$ to travel is $T$. So for $x$ position, $Tx/\lambda$ will be the time. Substituting it in (1), we get $$Y(x)=A\sin(2\pi x/\lambda)$$ $$Y(x)=A\sin(kx) \tag{2}$$
Can we find $$Y(x,t)=A\sin(kx+\phi-\omega t)$$ from a circle with radius as the amplitude $A$?
What I find difficult in doing it is the following at $t =0$ particle makes an angle $\phi$ (phase constant). Equation (2) is the displacement at time $t$.