If $\gamma$ is a circle and $\int_0^{2\pi}y(t) \, dt =0$ then $y(t)=a \sin t + b\cos t$

Question

I'm reading a paper on the isoperimetric theorem and I can't figure something out. Let $\gamma=(x(t),y(t))$, where $x(t)=-\int_0^t y(s) \, ds$, and $y(t)$ is a smooth function of period $2\pi$ such that $\int_0^{2\pi}y(t) \, dt=0$. It follows that $x(t)$ is also of period $2\pi$, so $\gamma$ is a smooth closed curve. It is then shown that $$x'(t)^2 + y'(t)^2=c^2 $$ where $c\in \mathbb{R}$ is a constant, and that $L=2\pi c$ where $L$ is the length of $\gamma$. It is also known that $\gamma$ must be a circumference and that $$\int_0^{2\pi}y'(t)^2 \, dt = \int_0^{2\pi} y(t)^2 \, dt$$ Take everything I said up until now as hypothesis. Now the bit I don't get: the author says that $\int_0^{2\pi}y(t) \, dt = 0$ easily implies that $y(t)=a \sin t + b \cos t$ where $a,b \in \mathbb{R}$. Why is this true?