Problem Solution

Weekly Problem 49 - 2012

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 $(1,0)$.