If-x-and-y-real-number-satisfy-x-5-2-y-12-2-196-then-the-minimum-value-of-x-2-y-2-is-

Question Number 96500 by bemath last updated on 02/Jun/20

$$\mathrm{If}\:{x}\:{and}\:{y}\:{real}\:{number}\:{satisfy} \\ $$$$\left({x}+\mathrm{5}\right)^{\mathrm{2}} +\left(\mathrm{y}−\mathrm{12}\right)^{\mathrm{2}} =\mathrm{196}\:,\:\mathrm{then}\: \\ $$$$\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:{is}\: \\ $$

Answered by bobhans last updated on 02/Jun/20

Let x+5 = 14 cos θ , y−12=14sin θ x^2 +y^2 = 365 + 28(12sin θ−5cos θ) = 365 + 364 sin (θ−β) where tan β=(5/(12)) so x^2 +y^2 has the minimum value 1 when θ = ((3π)/2) + arctan ((5/(12))) , i.e ⇒x = (5/(13)) & y = −((12)/(13))

$$\mathrm{Let}\:{x}+\mathrm{5}\:=\:\mathrm{14}\:\mathrm{cos}\:\theta\:,\:\mathrm{y}−\mathrm{12}=\mathrm{14sin}\:\theta \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:=\:\mathrm{365}\:+\:\mathrm{28}\left(\mathrm{12sin}\:\theta−\mathrm{5cos}\:\theta\right) \\ $$$$=\:\mathrm{365}\:+\:\mathrm{364}\:\mathrm{sin}\:\left(\theta−\beta\right)\:\mathrm{where}\:\mathrm{tan}\:\beta=\frac{\mathrm{5}}{\mathrm{12}} \\ $$$$\mathrm{so}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:{has}\:{the}\:{minimum}\:{value}\:\mathrm{1} \\ $$$${when}\:\theta\:=\:\frac{\mathrm{3}\pi}{\mathrm{2}}\:+\:\mathrm{arctan}\:\left(\frac{\mathrm{5}}{\mathrm{12}}\right)\:,\:\mathrm{i}.\mathrm{e} \\ $$$$\Rightarrow{x}\:=\:\frac{\mathrm{5}}{\mathrm{13}}\:\&\:\mathrm{y}\:=\:−\frac{\mathrm{12}}{\mathrm{13}} \\ $$

Commented by bemath last updated on 02/Jun/20

$$\mathrm{thanks} \\ $$

Commented by 1549442205 last updated on 02/Jun/20

I add a question:Which is the greatest value of the expression x^2 +y^2 ?

$$\mathrm{I}\:\mathrm{add}\:\mathrm{a}\:\mathrm{question}:\mathrm{Which}\:\mathrm{is}\:\mathrm{the}\:\mathrm{greatest}\: \\ $$$$\mathrm{value}\:\mathrm{of}\:\:\mathrm{the}\:\mathrm{expression}\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \:? \\ $$

Commented by MJS last updated on 02/Jun/20

$$\mathrm{365}+\mathrm{364}=\mathrm{729} \\ $$

Commented by bobhans last updated on 02/Jun/20

$$\mathrm{yes}=== \\ $$

Commented by bobhans last updated on 02/Jun/20

$$\mathrm{when}\:\theta\:=\:\frac{\pi}{\mathrm{2}}+\mathrm{arctan}\:\frac{\mathrm{5}}{\mathrm{12}}\: \\ $$

Commented by 1549442205 last updated on 02/Jun/20

you are all right!it is as following: we have x^2 +10x+25+y^2 −24y+144=196 ⇒x^2 +y^2 =27−10x+24y(1).Applyin Bunhiacopxky′s inequality we get (−10x+24y)^2 ≤(10^2 +24^2 )(x^2 +y^2 ) =676(x^2 +y^2 )⇒∣−10x+24y∣≤26(√(x^2 +y^2 )) (2) From (1),(2)we get x^2 +y^2 ≤27 +26(√(x^2 +y^2 ))⇔((√(x^2 +y^2 )) +1)((√(x^2 +y^2 )) −27)≤0 ⇒(√(x^2 +y^2 )) ≤27⇒x^2 +y^2 ≤729. The equality occurs if and only if { (((x/(−10))=(y/(24)))),((x^2 +y^2 =729)) :}⇔ { ((x=((−135)/(13)))),((y=((324)/(13)))) :}

$$\mathrm{you}\:\mathrm{are}\:\mathrm{all}\:\mathrm{right}!\mathrm{it}\:\mathrm{is}\:\mathrm{as}\:\mathrm{following}: \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{x}^{\mathrm{2}} +\mathrm{10x}+\mathrm{25}+\mathrm{y}^{\mathrm{2}} −\mathrm{24y}+\mathrm{144}=\mathrm{196} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{27}−\mathrm{10x}+\mathrm{24y}\left(\mathrm{1}\right).\mathrm{Applyin} \\ $$$$\mathrm{Bunhiacopxky}'\mathrm{s}\:\mathrm{inequality}\:\mathrm{we}\:\mathrm{get} \\ $$$$\left(−\mathrm{10x}+\mathrm{24y}\right)^{\mathrm{2}} \leqslant\left(\mathrm{10}^{\mathrm{2}} +\mathrm{24}^{\mathrm{2}} \right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right) \\ $$$$=\mathrm{676}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)\Rightarrow\mid−\mathrm{10x}+\mathrm{24y}\mid\leqslant\mathrm{26}\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }\:\left(\mathrm{2}\right) \\ $$$$\mathrm{From}\:\left(\mathrm{1}\right),\left(\mathrm{2}\right)\mathrm{we}\:\mathrm{get}\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \leqslant\mathrm{27} \\ $$$$+\mathrm{26}\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }\Leftrightarrow\left(\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }\:+\mathrm{1}\right)\left(\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }\:−\mathrm{27}\right)\leqslant\mathrm{0} \\ $$$$\Rightarrow\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \:}\:\leqslant\mathrm{27}\Rightarrow\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \leqslant\mathrm{729}. \\ $$$$\mathrm{The}\:\mathrm{equality}\:\mathrm{occurs}\:\mathrm{if}\:\mathrm{and}\:\mathrm{only} \\ $$$$\mathrm{if}\:\begin{cases}{\frac{\mathrm{x}}{−\mathrm{10}}=\frac{\mathrm{y}}{\mathrm{24}}}\\{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{729}}\end{cases}\Leftrightarrow\begin{cases}{\mathrm{x}=\frac{−\mathrm{135}}{\mathrm{13}}}\\{\mathrm{y}=\frac{\mathrm{324}}{\mathrm{13}}}\end{cases} \\ $$$$ \\ $$

Answered by john santu last updated on 02/Jun/20

i have a method in short if (x−a)^2 +(y−b)^2 =r^2 then x^2 +y^2 { ((max = {r+(√(a^2 +b^2 ))}^2 )),((min = {r−(√(a^2 +b^2 ))}^2 )) :}

$$\mathrm{i}\:\mathrm{have}\:\mathrm{a}\:\mathrm{method}\:\mathrm{in}\:\mathrm{short} \\ $$$$\mathrm{if}\:\left(\mathrm{x}−\mathrm{a}\right)^{\mathrm{2}} +\left(\mathrm{y}−\mathrm{b}\right)^{\mathrm{2}} =\mathrm{r}^{\mathrm{2}} \\ $$$$\mathrm{then}\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \:\:\begin{cases}{\mathrm{max}\:=\:\left\{\mathrm{r}+\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }\right\}^{\mathrm{2}} }\\{\mathrm{min}\:=\:\left\{\mathrm{r}−\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }\right\}^{\mathrm{2}} }\end{cases} \\ $$

Commented by bobhans last updated on 02/Jun/20

$$\mathrm{waw}====\mathrm{thanks}\: \\ $$

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