{l}{(x+2)^2+(y-1)^2=1}{z=my}{text{Solvefor}atext{where}}{a=z} ઉકેલો

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

x, y, z માટે ઉકેલો (જટિલ સમાધાન)

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

x, y, z માટે ઉકેલો

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