Question Number 210765 by shhhh last updated on 18/Aug/24
Answered by Berbere last updated on 19/Aug/24
$$\begin{cases}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{4}}\\{\left({x}−\mathrm{3}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{9}\Rightarrow−\mathrm{6}{x}=−\mathrm{4}}\end{cases} \\ $$$${x}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${y}^{\mathrm{2}} =\frac{\mathrm{32}}{\mathrm{9}} \\ $$$${z}=\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{1}\underset{−} {+}{i}\mathrm{2}\sqrt{\mathrm{2}}\right) \\ $$$$ \\ $$$$ \\ $$