Question Number 156520 by aliyn last updated on 12/Oct/21
$$\boldsymbol{{if}}\:\mid\:\boldsymbol{{z}}+\mathrm{6}\boldsymbol{{i}}\:\mid\leqslant\mathrm{5}\:\boldsymbol{{find}}\:\boldsymbol{{the}}\:\boldsymbol{{maximum}}\:\boldsymbol{{and}}\:\boldsymbol{{minimum}}\:\boldsymbol{{value}}\:\boldsymbol{{of}}\:\mid\:\boldsymbol{{z}}+\mathrm{5}\mid\:? \\ $$
Answered by TheSupreme last updated on 12/Oct/21
$${x}^{\mathrm{2}} +\left(\mathrm{6}+{y}\right)^{\mathrm{2}} \leqslant\mathrm{5}\:\left({circle}\:{center}\:−\mathrm{6}{i}\:{radius}\:\sqrt{\left.\mathrm{5}\right)}\right. \\ $$$$\left({x}+\mathrm{5}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\theta\left({x},{y}\right)\:\left({circle}\:{of}\:{center}\:−\mathrm{5}\:{radius}\:\sqrt{\theta}\right) \\ $$$$\sqrt{\theta}+\sqrt{\mathrm{5}}=\sqrt{\mathrm{5}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} } \\ $$$$\sqrt{\theta}−\sqrt{\mathrm{5}}=\sqrt{\mathrm{5}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} } \\ $$$${min}=\sqrt{\mathrm{61}}−\sqrt{\mathrm{5}} \\ $$$${max}=\sqrt{\mathrm{61}}+\sqrt{\mathrm{5}} \\ $$