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Question Number 208023 by efronzo1 last updated on 02/Jun/24

Answered by mr W last updated on 02/Jun/24

$$x=a+\mid r\mid \cos \theta$$$$\Rightarrow y=b+\mid r\mid \sin \theta$$$$\sqrt{x^2+y^2}$$$$=\sqrt{a^2+b^2+r^2+2\mid r\mid (a\cos \theta +b\sin \theta )}$$$$=\sqrt{a^2+b^2+r^2+2\mid r\mid \sqrt{a^2+b^2}\cos (\theta -{\tan }^{-1}\frac{b}{a})}$$$$\Rightarrow {\left(\sqrt{x^2+y^2}\right)}_{max}=\sqrt{a^2+b^2+r^2+2\mid r\mid \sqrt{a^2+b^2}}$$$$=\sqrt{a^2+b^2}+\mid r\mid$$$$\Rightarrow {\left(\sqrt{x^2+y^2}\right)}_{min}=\sqrt{a^2+b^2+r^2-2\mid r\mid \sqrt{a^2+b^2}}$$$$=\mid \sqrt{a^2+b^2}-\mid r\mid \mid$$

Commented by efronzo1 last updated on 02/Jun/24

$$\mathrm{when}\cancel{\underset{―}{⋖}}\ne 0,\mathrm{if}⋖\cancel{}$$

Commented by mr W last updated on 02/Jun/24

$${it}^{\prime }svalidforanya,b\in R.$$

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