Question Number 86063 by john santu last updated on 26/Mar/20
Answered by mr W last updated on 26/Mar/20
$$\left(\frac{\mathrm{1}}{{R}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{9}}\right)^{\mathrm{2}} =\mathrm{2}\left(\frac{\mathrm{1}}{{R}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{9}^{\mathrm{2}} }\right) \\ $$$$\Rightarrow{R}=\frac{\mathrm{18}}{\mathrm{7}} \\ $$
Commented by jagoll last updated on 26/Mar/20
$$\mathrm{how}\:\mathrm{sir}\:,\:\mathrm{got}\:\mathrm{it}\:?\: \\ $$$$\mathrm{explain}\:\mathrm{me}.\:\mathrm{please} \\ $$$$ \\ $$
Commented by jagoll last updated on 27/Mar/20
$$\mathrm{sir}\:\mathrm{it}\:\mathrm{not}\: \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{R}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}}+\:\frac{\mathrm{1}}{\mathrm{9}}\right)^{\mathrm{2}} \:=\: \\ $$$$\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{R}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{9}^{\mathrm{2}} }\right) \\ $$$$\mathrm{Descartes}\:\mathrm{theorem} \\ $$
Commented by mr W last updated on 27/Mar/20
$${see}\:{Q}\mathrm{77681} \\ $$
Commented by mr W last updated on 27/Mar/20
$${the}\:{big}\:{circle}\:{is}\:{touched}\:{on}\:{its}\:{inside}, \\ $$$${therefore}\:{its}\:{curvature}\:{is}\:{negative}! \\ $$$${otherwise}\:{you}\:{won}'{t}\:{be}\:{able}\:{to}\:{get}\:{a} \\ $$$${solution}. \\ $$
Commented by jagoll last updated on 27/Mar/20
$$\mathrm{ok}\:\mathrm{sir}.\:\mathrm{thank}\:\mathrm{you} \\ $$
Commented by john santu last updated on 27/Mar/20
$${thank}\:{you} \\ $$