Question Number 213818 by ajfour last updated on 17/Nov/24
Commented by ajfour last updated on 17/Nov/24
$${Find}\:{R}\:{in}\:{terms}\:{smaller}\:{radii}\:{a},\:{b}. \\ $$
Answered by mr W last updated on 17/Nov/24
Commented by ajfour last updated on 17/Nov/24
$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\left({x}−{R}−{r}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${slope}\:{of}\:{common}\:{tangent} \\ $$$${m}^{\mathrm{2}} =\frac{\left({R}−{a}\right)^{\mathrm{2}} }{\mathrm{4}{Ra}}=\frac{\mathrm{4}{Rb}}{\left({R}−{b}\right)^{\mathrm{2}} } \\ $$$${R}^{\mathrm{2}} −\left({a}+{b}\right){R}+{ab}=\mathrm{4}{R}\sqrt{{ab}} \\ $$$${R}=\frac{{a}+{b}}{\mathrm{2}}+\mathrm{2}\sqrt{{ab}}+\sqrt{\left(\frac{{a}+{b}}{\mathrm{2}}+\mathrm{2}\sqrt{{ab}}\right)^{\mathrm{2}} −{ab}} \\ $$$${say}\:{b}=\mathrm{1},\:{a}=\mathrm{2} \\ $$$${R}=\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{2}\sqrt{\mathrm{2}}+\sqrt{\frac{\mathrm{9}}{\mathrm{4}}+\mathrm{8}+\mathrm{6}\sqrt{\mathrm{2}}−\mathrm{2}} \\ $$$$\:\:\:\:{R}=\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{2}\sqrt{\mathrm{2}}+\sqrt{\frac{\mathrm{33}}{\mathrm{4}}+\mathrm{6}\sqrt{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:\approx\:\mathrm{8}.\mathrm{4193} \\ $$
Commented by mr W last updated on 17/Nov/24
$${yes}.\:{now}\:{we}\:{got}\:{the}\:{same}. \\ $$
Commented by mr W last updated on 17/Nov/24
$$\sqrt{\left({R}+{a}\right)^{\mathrm{2}} −\left({R}−{a}\right)^{\mathrm{2}} }+\sqrt{\left({R}+{b}\right)^{\mathrm{2}} −\left({R}−{b}\right)^{\mathrm{2}} }=\sqrt{\left({R}+{a}\right)^{\mathrm{2}} +\left({R}+{b}\right)^{\mathrm{2}} −\left({a}−{b}\right)^{\mathrm{2}} } \\ $$$$\sqrt{\mathrm{2}{aR}}+\sqrt{\mathrm{2}{bR}}=\sqrt{\left({R}+{a}\right)\left({R}+{b}\right)} \\ $$$${R}^{\mathrm{2}} −\left({a}+{b}+\mathrm{4}\sqrt{{ab}}\right){R}+{ab}=\mathrm{0} \\ $$$${R}=\frac{{a}+{b}+\mathrm{4}\sqrt{{ab}}+\sqrt{\left({a}+{b}+\mathrm{6}\sqrt{{ab}}\right)\left({a}+{b}+\mathrm{2}\sqrt{{ab}}\right)}}{\mathrm{2}} \\ $$
Commented by ajfour last updated on 17/Nov/24
$${plz}\:{explain}\:{sir},\:{R}.{H}.{S}.\:{of}\:{first}\:{line}.. \\ $$
Commented by mr W last updated on 17/Nov/24
$${AB}^{\mathrm{2}} =\left(\sqrt{\left({R}+{a}\right)^{\mathrm{2}} −\left({R}−{a}\right)^{\mathrm{2}} }+\sqrt{\left({R}+{b}\right)^{\mathrm{2}} −\left({R}−{b}\right)^{\mathrm{2}} }\right)^{\mathrm{2}} +\left({a}−{b}\right)^{\mathrm{2}} \\ $$$${AB}^{\mathrm{2}} =\left({R}+{a}\right)^{\mathrm{2}} +\left({R}+{b}\right)^{\mathrm{2}} \\ $$
Commented by mr W last updated on 17/Nov/24
Commented by ajfour last updated on 17/Nov/24
