Menu Close

a-gt-0-b-gt-0-x-1-2-y-7-2-a-2-x-2-2-y-3-2-b-2-Find-a-b-min-




Question Number 183380 by Shrinava last updated on 25/Dec/22
a>0 , b>0   { (((x−1)^2  + (y−7)^2  = a^2 )),(((x−2)^2  + (y−3)^2  = b^2 )) :}  Find:   (a+b)_(min)  = ?
$$\mathrm{a}>\mathrm{0}\:,\:\mathrm{b}>\mathrm{0} \\ $$$$\begin{cases}{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} \:+\:\left(\mathrm{y}−\mathrm{7}\right)^{\mathrm{2}} \:=\:\mathrm{a}^{\mathrm{2}} }\\{\left(\mathrm{x}−\mathrm{2}\right)^{\mathrm{2}} \:+\:\left(\mathrm{y}−\mathrm{3}\right)^{\mathrm{2}} \:=\:\mathrm{b}^{\mathrm{2}} }\end{cases} \\ $$$$\mathrm{Find}:\:\:\:\left(\mathrm{a}+\mathrm{b}\right)_{\boldsymbol{\mathrm{min}}} \:=\:? \\ $$
Commented by Frix last updated on 25/Dec/22
(√(17))= the distance of the centers  If we see it as two independent circles  obviously the limit of the minimum is 0
$$\sqrt{\mathrm{17}}=\:\mathrm{the}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{the}\:\mathrm{centers} \\ $$$$\mathrm{If}\:\mathrm{we}\:\mathrm{see}\:\mathrm{it}\:\mathrm{as}\:\mathrm{two}\:\mathrm{independent}\:\mathrm{circles} \\ $$$$\mathrm{obviously}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{of}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{is}\:\mathrm{0} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *