Question Number 208023 by efronzo1 last updated on 02/Jun/24
Answered by mr W last updated on 02/Jun/24
$${x}={a}+\mid{r}\mid\:\mathrm{cos}\:\theta \\ $$$$\Rightarrow{y}={b}+\mid{r}\mid\:\mathrm{sin}\:\theta \\ $$$$\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} } \\ $$$$=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{2}\mid{r}\mid\left({a}\:\mathrm{cos}\:\theta+{b}\:\mathrm{sin}\:\theta\right)} \\ $$$$=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{2}\mid{r}\mid\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:\mathrm{cos}\:\left(\theta−\mathrm{tan}^{−\mathrm{1}} \frac{{b}}{{a}}\right)} \\ $$$$\Rightarrow\left(\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\right)_{{max}} =\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{2}\mid{r}\mid\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }+\mid{r}\mid \\ $$$$\Rightarrow\left(\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\right)_{{min}} =\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}\mid{r}\mid\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mid\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }−\mid{r}\mid\mid \\ $$
Commented by efronzo1 last updated on 02/Jun/24
$$\mathrm{when}\:\:\cancel{\underline{\lessdot}}\neq\:\mathrm{0}\:,\:\mathrm{if}\:\:\lessdot\cancel{ } \\ $$
Commented by mr W last updated on 02/Jun/24
$${it}'{s}\:{valid}\:{for}\:{any}\:{a},\:{b}\in{R}. \\ $$