Selesaikan x^2+(y-sqrt{x^2})^2=1

$$x ^ { 2 } + ( y - \sqrt { x ^ { 2 } } ) ^ { 2 } = 1$$

Cari nilai x

$$\left\{\begin{matrix}x=\frac{\sqrt{-2y\sqrt{2-y^{2}}+2}}{2}\text{; }x=-\frac{\sqrt{-2y\sqrt{2-y^{2}}+2}}{2}\text{, }&y=0\text{ or }\left(y\leq \sqrt{2}\text{ and }y\geq 1\text{ and }y\sqrt{2-y^{2}}\leq 1\right)\\x=\frac{\sqrt{2y\sqrt{2-y^{2}}+2}}{2}\text{; }x=-\frac{\sqrt{2y\sqrt{2-y^{2}}+2}}{2}\text{, }&\left(y\sqrt{2-y^{2}}\geq -1\text{ and }y\leq 0\text{ and }-y\sqrt{2-y^{2}}\leq 1\text{ and }y\geq -1\right)\text{ or }\left(y\geq 0\text{ and }y\leq 1\text{ and }y\sqrt{2-y^{2}}\leq 1\right)\text{ or }\left(y\leq \sqrt{2}\text{ and }y\geq 1\text{ and }y\sqrt{2-y^{2}}\leq 1\right)\end{matrix}\right,$$

Cari nilai y

$$y=-\sqrt{1-x^{2}}+|x|$$

$$y=\sqrt{1-x^{2}}+|x|\text{, }|x|\leq 1$$