Question Number 212723 by ajfour last updated on 22/Oct/24
Commented by ajfour last updated on 22/Oct/24
$${Find}\:\:{R}_{{min}} \\ $$
Answered by mr W last updated on 22/Oct/24
Commented by mr W last updated on 25/Oct/24
![OE=x, say OC=(√((R−a)^2 −a^2 ))=(√(R(R−2a))) OD=(√((R−b)^2 −b^2 ))=(√(R(R−2b))) CE=(√(R(R−2a)))−x DE=(√(R(R−2b)))+x CD=(√(R(R−2a)))+(√(R(R−2b))) AB^2 =(b−a)^2 +[(√(R(R−2a)))+(√(R(R−2b)))]^2 AB^2 =(a+b)^2 +[(√(R(R−2b)))+x−(√(R(R−2a)))+x]^2 (a+b)^2 +[2x+(√(R(R−2b)))−(√(R(R−2a)))]^2 =(b−a)^2 +[(√(R(R−2a)))+(√(R(R−2b)))]^2 ab=[(√(R(R−2b)))+x][(√(R(R−2a)))−x] (dR/dx)=0 (√(R(R−2b)))+x=(√(R(R−2a)))−x ⇒x=((√(R(R−2a)))−(√(R(R−2b))))/2 4ab=((√(R(R−2a)))+(√(R(R−2b))))^2 2ab+(a+b)R−R^2 =R(√((R−2a)(R−2b))))^2 (6ab−a^2 −b^2 )R^2 −4ab(a+b)R−4a^2 b^2 =0 ⇒R_(min) =((2ab(2(√(2ab))+a+b))/(6ab−a^2 −b^2 )) ✓ =((2ab(2(√(2ab))+a+b))/(8ab−(a+b)^2 )) =((2ab(2(√(2ab))+a+b))/((2(√(2ab))+a+b)(2(√(2ab))−a−b))) =((2ab)/(2(√(2ab))−a−b)) ✓](https://www.tinkutara.com/question/Q212727.png)
$${OE}={x},\:{say} \\ $$$${OC}=\sqrt{\left({R}−{a}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} }=\sqrt{{R}\left({R}−\mathrm{2}{a}\right)} \\ $$$${OD}=\sqrt{\left({R}−{b}\right)^{\mathrm{2}} −{b}^{\mathrm{2}} }=\sqrt{{R}\left({R}−\mathrm{2}{b}\right)} \\ $$$${CE}=\sqrt{{R}\left({R}−\mathrm{2}{a}\right)}−{x} \\ $$$${DE}=\sqrt{{R}\left({R}−\mathrm{2}{b}\right)}+{x} \\ $$$${CD}=\sqrt{{R}\left({R}−\mathrm{2}{a}\right)}+\sqrt{{R}\left({R}−\mathrm{2}{b}\right)} \\ $$$${AB}^{\mathrm{2}} =\left({b}−{a}\right)^{\mathrm{2}} +\left[\sqrt{{R}\left({R}−\mathrm{2}{a}\right)}+\sqrt{{R}\left({R}−\mathrm{2}{b}\right)}\right]^{\mathrm{2}} \\ $$$${AB}^{\mathrm{2}} =\left({a}+{b}\right)^{\mathrm{2}} +\left[\sqrt{{R}\left({R}−\mathrm{2}{b}\right)}+{x}−\sqrt{{R}\left({R}−\mathrm{2}{a}\right)}+{x}\right]^{\mathrm{2}} \\ $$$$\left({a}+{b}\right)^{\mathrm{2}} +\left[\mathrm{2}{x}+\sqrt{{R}\left({R}−\mathrm{2}{b}\right)}−\sqrt{{R}\left({R}−\mathrm{2}{a}\right)}\right]^{\mathrm{2}} =\left({b}−{a}\right)^{\mathrm{2}} +\left[\sqrt{{R}\left({R}−\mathrm{2}{a}\right)}+\sqrt{{R}\left({R}−\mathrm{2}{b}\right)}\right]^{\mathrm{2}} \\ $$$${ab}=\left[\sqrt{{R}\left({R}−\mathrm{2}{b}\right)}+{x}\right]\left[\sqrt{{R}\left({R}−\mathrm{2}{a}\right)}−{x}\right] \\ $$$$\frac{{dR}}{{dx}}=\mathrm{0} \\ $$$$\sqrt{{R}\left({R}−\mathrm{2}{b}\right)}+{x}=\sqrt{{R}\left({R}−\mathrm{2}{a}\right)}−{x} \\ $$$$\Rightarrow{x}=\left(\sqrt{{R}\left({R}−\mathrm{2}{a}\right)}−\sqrt{{R}\left({R}−\mathrm{2}{b}\right)}\right)/\mathrm{2} \\ $$$$\mathrm{4}{ab}=\left(\sqrt{{R}\left({R}−\mathrm{2}{a}\right)}+\sqrt{{R}\left({R}−\mathrm{2}{b}\right)}\right)^{\mathrm{2}} \\ $$$$\left.\mathrm{2}{ab}+\left({a}+{b}\right){R}−{R}^{\mathrm{2}} ={R}\sqrt{\left({R}−\mathrm{2}{a}\right)\left({R}−\mathrm{2}{b}\right)}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{6}{ab}−{a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){R}^{\mathrm{2}} −\mathrm{4}{ab}\left({a}+{b}\right){R}−\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{R}_{{min}} =\frac{\mathrm{2}{ab}\left(\mathrm{2}\sqrt{\mathrm{2}{ab}}+{a}+{b}\right)}{\mathrm{6}{ab}−{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\:\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}{ab}\left(\mathrm{2}\sqrt{\mathrm{2}{ab}}+{a}+{b}\right)}{\mathrm{8}{ab}−\left({a}+{b}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}{ab}\left(\mathrm{2}\sqrt{\mathrm{2}{ab}}+{a}+{b}\right)}{\left(\mathrm{2}\sqrt{\mathrm{2}{ab}}+{a}+{b}\right)\left(\mathrm{2}\sqrt{\mathrm{2}{ab}}−{a}−{b}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}{ab}}{\mathrm{2}\sqrt{\mathrm{2}{ab}}−{a}−{b}}\:\checkmark\:\:\: \\ $$
Commented by ajfour last updated on 25/Oct/24
$${could}\:{this}\:{be}\:{true}\:{instead}\: \\ $$$$\:\:{R}_{{min}} =\frac{\mathrm{2}{ab}}{\mathrm{2}\sqrt{\mathrm{2}{ab}}−\left({a}+{b}\right)} \\ $$$${say}\:{if}\:{a}=\mathrm{2},\:{b}=\mathrm{1} \\ $$$${R}_{{min}} =\frac{\mathrm{4}×\mathrm{7}}{\mathrm{7}}=\mathrm{4} \\ $$$${My}\:{ans}\:{gives}\:{R}_{{min}} =\frac{\mathrm{4}}{\mathrm{4}−\left(\mathrm{3}\right)}=\mathrm{4} \\ $$$${another}\:\:\:{a}=\mathrm{5},\:{b}=\mathrm{2} \\ $$$${R}_{{min}} =\frac{\mathrm{20}\left(\mathrm{4}\sqrt{\mathrm{5}}+\mathrm{7}\right)}{\left(\mathrm{60}−\mathrm{29}\right)}=\frac{\mathrm{20}\left(\mathrm{4}\sqrt{\mathrm{5}}+\mathrm{7}\right)}{\mathrm{31}} \\ $$$${My}\:{ans}\:\:{R}_{{min}} =\frac{\mathrm{20}}{\mathrm{4}\sqrt{\mathrm{5}}−\mathrm{7}}=\frac{\mathrm{20}\left(\mathrm{4}\sqrt{\mathrm{5}}+\mathrm{7}\right)}{\mathrm{31}} \\ $$$$\:\:\Rightarrow\:{my}\:{answer}\:{msy}\:{be}\:{equal}\:{to}\:{yours} \\ $$$${i}\:{took}\:{R}_{{min}\:} \:{occurs}\:{when}\:{both} \\ $$$${circles}\:{touch}\:{the}\:{line}\:{st}\:{same}\:{point}. \\ $$
Commented by mr W last updated on 25/Oct/24
$${yes},\:{they}\:{are}\:{the}\:{same}. \\ $$