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Question Number 156520 by aliyn last updated on 12/Oct/21

$$\mathit{i}\mathit{f}\mid \mathit{z}+6\mathit{i}\mid ⩽5\mathit{f}\mathit{i}\mathit{n}\mathit{d}\mathit{t}\mathit{h}\mathit{e}\mathit{m}\mathit{a}\mathit{x}\mathit{i}\mathit{m}\mathit{u}\mathit{m}\mathit{a}\mathit{n}\mathit{d}\mathit{m}\mathit{i}\mathit{n}\mathit{i}\mathit{m}\mathit{u}\mathit{m}\mathit{v}\mathit{a}\mathit{l}\mathit{u}\mathit{e}\mathit{o}\mathit{f}\mid \mathit{z}+5\mid ?$$

Answered by TheSupreme last updated on 12/Oct/21

$$x^2+{(6+y)}^2⩽5(circlecenter-6iradius\sqrt{5)}$$$${(x+5)}^2+y^2=\theta (x,y)(circleofcenter-5radius\sqrt{\theta })$$$$\sqrt{\theta }+\sqrt{5}=\sqrt{5^2+6^2}$$$$\sqrt{\theta }-\sqrt{5}=\sqrt{5^2+6^2}$$$$min=\sqrt{61}-\sqrt{5}$$$$max=\sqrt{61}+\sqrt{5}$$

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