समाधान x+y/x^2+y^2=1

$$\frac { x + y } { x ^ { 2 } + y ^ { 2 } } = 1$$

x को लागि हल गर्नुहोस् (complex solution)

$$\left\{\begin{matrix}\\x=\frac{\sqrt{1+4y-4y^{2}}+1}{2}\text{, }&\text{unconditionally}\\x=\frac{-\sqrt{1+4y-4y^{2}}+1}{2}\text{, }&y\neq 0\end{matrix}\right.$$

y को लागि हल गर्नुहोस् (complex solution)

$$\left\{\begin{matrix}\\y=\frac{\sqrt{1+4x-4x^{2}}+1}{2}\text{, }&\text{unconditionally}\\y=\frac{-\sqrt{1+4x-4x^{2}}+1}{2}\text{, }&x\neq 0\end{matrix}\right.$$

x को लागि हल गर्नुहोस्

$$\left\{\begin{matrix}x=\frac{\sqrt{1+4y-4y^{2}}+1}{2}\text{, }&y\geq \frac{1-\sqrt{2}}{2}\text{ and }y\leq \frac{\sqrt{2}+1}{2}\\x=\frac{-\sqrt{1+4y-4y^{2}}+1}{2}\text{, }&y\geq \frac{1-\sqrt{2}}{2}\text{ and }y\leq \frac{\sqrt{2}+1}{2}\text{ and }y\neq 0\end{matrix}\right.$$

y को लागि हल गर्नुहोस्

$$\left\{\begin{matrix}y=\frac{\sqrt{1+4x-4x^{2}}+1}{2}\text{, }&x\geq \frac{1-\sqrt{2}}{2}\text{ and }x\leq \frac{\sqrt{2}+1}{2}\\y=\frac{-\sqrt{1+4x-4x^{2}}+1}{2}\text{, }&x\geq \frac{1-\sqrt{2}}{2}\text{ and }x\leq \frac{\sqrt{2}+1}{2}\text{ and }x\neq 0\end{matrix}\right.$$