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Question Number 213818 by ajfour last updated on 17/Nov/24

Commented by ajfour last updated on 17/Nov/24

Find R in terms smaller radii a, b.

$${Find}\:{R}\:{in}\:{terms}\:{smaller}\:{radii}\:{a},\:{b}. \\ $$

Answered by mr W last updated on 17/Nov/24

Commented by ajfour last updated on 17/Nov/24

x^2 +y^2 =R^2 (x−R−r)^2 +y^2 =r^2 slope of common tangent m^2 =(((R−a)^2 )/(4Ra))=((4Rb)/((R−b)^2 )) R^2 −(a+b)R+ab=4R(√(ab)) R=((a+b)/2)+2(√(ab))+(√((((a+b)/2)+2(√(ab)))^2 −ab)) say b=1, a=2 R=(3/2)+2(√2)+(√((9/4)+8+6(√2)−2)) R=(3/2)+2(√2)+(√(((33)/4)+6(√2))) ≈ 8.4193

$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\left({x}−{R}−{r}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${slope}\:{of}\:{common}\:{tangent} \\ $$$${m}^{\mathrm{2}} =\frac{\left({R}−{a}\right)^{\mathrm{2}} }{\mathrm{4}{Ra}}=\frac{\mathrm{4}{Rb}}{\left({R}−{b}\right)^{\mathrm{2}} } \\ $$$${R}^{\mathrm{2}} −\left({a}+{b}\right){R}+{ab}=\mathrm{4}{R}\sqrt{{ab}} \\ $$$${R}=\frac{{a}+{b}}{\mathrm{2}}+\mathrm{2}\sqrt{{ab}}+\sqrt{\left(\frac{{a}+{b}}{\mathrm{2}}+\mathrm{2}\sqrt{{ab}}\right)^{\mathrm{2}} −{ab}} \\ $$$${say}\:{b}=\mathrm{1},\:{a}=\mathrm{2} \\ $$$${R}=\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{2}\sqrt{\mathrm{2}}+\sqrt{\frac{\mathrm{9}}{\mathrm{4}}+\mathrm{8}+\mathrm{6}\sqrt{\mathrm{2}}−\mathrm{2}} \\ $$$$\:\:\:\:{R}=\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{2}\sqrt{\mathrm{2}}+\sqrt{\frac{\mathrm{33}}{\mathrm{4}}+\mathrm{6}\sqrt{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:\approx\:\mathrm{8}.\mathrm{4193} \\ $$

Commented by mr W last updated on 17/Nov/24

yes. now we got the same.

$${yes}.\:{now}\:{we}\:{got}\:{the}\:{same}. \\ $$

Commented by mr W last updated on 18/Nov/24

(√((R+a)^2 −(R−a)^2 ))+(√((R+b)^2 −(R−b)^2 ))=(√((R+a)^2 +(R+b)^2 −(a−b)^2 )) (√(2aR))+(√(2bR))=(√((R+a)(R+b))) R^2 −(a+b+4(√(ab)))R+ab=0 R=((a+b)/2)+2(√(ab))+(√((((a+b)/2)+(√(ab)))(((a+b)/2)+3(√(ab)))))

$$\sqrt{\left({R}+{a}\right)^{\mathrm{2}} −\left({R}−{a}\right)^{\mathrm{2}} }+\sqrt{\left({R}+{b}\right)^{\mathrm{2}} −\left({R}−{b}\right)^{\mathrm{2}} }=\sqrt{\left({R}+{a}\right)^{\mathrm{2}} +\left({R}+{b}\right)^{\mathrm{2}} −\left({a}−{b}\right)^{\mathrm{2}} } \\ $$$$\sqrt{\mathrm{2}{aR}}+\sqrt{\mathrm{2}{bR}}=\sqrt{\left({R}+{a}\right)\left({R}+{b}\right)} \\ $$$${R}^{\mathrm{2}} −\left({a}+{b}+\mathrm{4}\sqrt{{ab}}\right){R}+{ab}=\mathrm{0} \\ $$$${R}=\frac{{a}+{b}}{\mathrm{2}}+\mathrm{2}\sqrt{{ab}}+\sqrt{\left(\frac{{a}+{b}}{\mathrm{2}}+\sqrt{{ab}}\right)\left(\frac{{a}+{b}}{\mathrm{2}}+\mathrm{3}\sqrt{{ab}}\right)} \\ $$

Commented by ajfour last updated on 17/Nov/24

plz explain sir, R.H.S. of first line..

$${plz}\:{explain}\:{sir},\:{R}.{H}.{S}.\:{of}\:{first}\:{line}.. \\ $$

Commented by mr W last updated on 17/Nov/24

AB^2 =((√((R+a)^2 −(R−a)^2 ))+(√((R+b)^2 −(R−b)^2 )))^2 +(a−b)^2 AB^2 =(R+a)^2 +(R+b)^2

$${AB}^{\mathrm{2}} =\left(\sqrt{\left({R}+{a}\right)^{\mathrm{2}} −\left({R}−{a}\right)^{\mathrm{2}} }+\sqrt{\left({R}+{b}\right)^{\mathrm{2}} −\left({R}−{b}\right)^{\mathrm{2}} }\right)^{\mathrm{2}} +\left({a}−{b}\right)^{\mathrm{2}} \\ $$$${AB}^{\mathrm{2}} =\left({R}+{a}\right)^{\mathrm{2}} +\left({R}+{b}\right)^{\mathrm{2}} \\ $$

Commented by mr W last updated on 17/Nov/24

Commented by ajfour last updated on 17/Nov/24

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