समाधान {l}{(x+2)^2+(y-1)^2=1}{z=my}{a=z}{text{Solvefor}btext{where}}{b=a}

$$\left. \begin{array} { l } { {(x + 2)} ^ {2} + {(y - 1)} ^ {2} = 1 }\\ { z = m y }\\ { a = z }\\ { \text{Solve for } b \text{ where} } \\ { b = a } \end{array} \right.$$

x, y, z को लागि हल गर्नुहोस् (complex solution)

$$\left\{\begin{matrix}x=\frac{i\sqrt{a}\sqrt{a-2m}-2m}{m}\text{, }y=\frac{a}{m}\text{, }z=a\text{; }x=\frac{-i\sqrt{a}\sqrt{a-2m}-2m}{m}\text{, }y=\frac{a}{m}\text{, }z=a\text{, }&m\neq 0\text{ and }a=b\\x\in \mathrm{C}\text{, }y=\sqrt{-\left(x+1\right)\left(x+3\right)}+1\text{, }z=0\text{; }x\in \mathrm{C}\text{, }y=-\sqrt{-\left(x+1\right)\left(x+3\right)}+1\text{, }z=0\text{, }&a=0\text{ and }m=0\text{ and }b=0\end{matrix}\right.$$

x, y, z को लागि हल गर्नुहोस्

$$\left\{\begin{matrix}x=-\frac{\sqrt{b\left(2m-b\right)}+2|m|}{|m|}\text{, }y=\frac{b}{m}\text{, }z=b\text{; }x=\frac{\sqrt{b\left(2m-b\right)}-2|m|}{|m|}\text{, }y=\frac{b}{m}\text{, }z=b\text{, }&\left(m\geq \frac{b}{2}\text{ and }b>0\text{ and }a=b\right)\text{ or }\left(m=\frac{b}{2}\text{ and }b\neq 0\text{ and }a=b\right)\text{ or }\left(b<0\text{ and }m\leq \frac{b}{2}\text{ and }a=b\right)\\x\in \begin{bmatrix}-3,-1\end{bmatrix}\text{, }y=\sqrt{-\left(x+1\right)\left(x+3\right)}+1\text{, }z=0\text{; }x\in \begin{bmatrix}-3,-1\end{bmatrix}\text{, }y=-\sqrt{-\left(x+1\right)\left(x+3\right)}+1\text{, }z=0\text{, }&a=0\text{ and }m=0\text{ and }b=0\end{matrix}\right.$$