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Question Number 176501 by infinityaction last updated on 20/Sep/22

$$\mathrm{find}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{x}+\mathrm{y}\mathrm{such}\mathrm{that}$$$${(x-2)}^2+{(y-4)}^2=49$$

Commented by cortano1 last updated on 20/Sep/22

$$\mathrm{let}\mathrm{x}+\mathrm{y}=\mathrm{k}\Rightarrow \mathrm{x}+\mathrm{y}-\mathrm{k}=0$$$$\mathrm{is}\mathrm{tangent}\mathrm{to}\mathrm{circle}$$$$\mathrm{so}7=\frac{\mid 2+4-\mathrm{k}\mid }{\sqrt{2}}$$$$\Rightarrow \mid \mathrm{k}-6\mid =7\sqrt{2}$$$$\Rightarrow -7\sqrt{2}⩽\mathrm{k}-6⩽7\sqrt{2}$$$$\Rightarrow 6-7\sqrt{2}⩽\mathrm{k}⩽6+7\sqrt{2}$$$$\mathrm{Therefore}\mathrm{range}\mathrm{of}\mathrm{x}+\mathrm{y}$$$$\mathrm{is}[6-7\sqrt{2},6+7\sqrt{2}]$$

Answered by Peace last updated on 20/Sep/22

$$\{\begin{array}{l}x=2+7cos\left(t\right)\\ y=4+7sin\left(t\right)\end{array}t\in [0,2\pi [$$$$x+y=6+7(sin\left(t\right)+cos\left(t\right))=6+7\sqrt{2}\left(sin(t+\frac{\pi }{4})\right)$$$$x+y\in [6-7\sqrt{2},6+7\sqrt{2}]$$

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