Three, (Full marks 20 points) On the circle $C_{:}(x-1)^{2}+y^{2}=2$ there are two moving points $A$ and $B$, and they satisfy the condition $\angle A) B=90^{\circ}(O$ being the origin). Find the equation of the trajectory of point $P$ of the parallelogram $O A P B$ with $O A, O B$ as adjacent sides.
Three, let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), P(x, y)$. According to the problem, we have
$$
\begin{array}{l}
\left(x_{1}-1\right)^{2}+y_{1}^{2}=2, \\
\left(x_{2}-1\right)^{2}+y_{2}^{2}=2 .
\end{array}
$$

By adding (1) and (2), we get
$$
x_{1}^{2}+y_{1}^{2}+x_{2}^{2}+y_{2}^{2}-2\left(x_{1}+x_{2}\right)=2 .
$$

Since $O A P B$ is a parallelogram, we have
$$
\begin{array}{l}
x_{1}^{2}+y_{1}^{2}+x_{2}^{2}+y_{2}^{2}=x^{2}+y^{2}, \\
x_{1}+x_{2}=x .
\end{array}
$$

Substituting (4) and (5) into (3), we get
$$
(x-1)^{2}+y^{2}=(\sqrt{3})^{2} \text {. }
$$

Therefore, the locus of point $P$ is a circle with center $(1,0)$ and radius $\sqrt{3}$.