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We have $y^{2}=x(2-x)$.
$y^{2}\ge0$ for all real $y$, hence $x(2-x)\ge0$.
Hence $0\le x\le2$.
In fact we can rewrite the equation as $(x-1)^{2}+y^{2}=1$
So this is a circle of radius $1$ with centre at $(1,0)$.