Menu Close

Question-86063




Question Number 86063 by john santu last updated on 26/Mar/20
Answered by mr W last updated on 26/Mar/20
((1/R)+(1/3)+(1/6)−(1/9))^2 =2((1/R^2 )+(1/3^2 )+(1/6^2 )+(1/9^2 ))  ⇒R=((18)/7)
$$\left(\frac{\mathrm{1}}{{R}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{9}}\right)^{\mathrm{2}} =\mathrm{2}\left(\frac{\mathrm{1}}{{R}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{9}^{\mathrm{2}} }\right) \\ $$$$\Rightarrow{R}=\frac{\mathrm{18}}{\mathrm{7}} \\ $$
Commented by jagoll last updated on 26/Mar/20
how sir , got it ?   explain me. please
$$\mathrm{how}\:\mathrm{sir}\:,\:\mathrm{got}\:\mathrm{it}\:?\: \\ $$$$\mathrm{explain}\:\mathrm{me}.\:\mathrm{please} \\ $$$$ \\ $$
Commented by jagoll last updated on 27/Mar/20
sir it not   ((1/R)+(1/3)+(1/6)+ (1/9))^2  =   2((1/R^2 )+(1/3^2 )+(1/6^2 )+(1/9^2 ))  Descartes theorem
$$\mathrm{sir}\:\mathrm{it}\:\mathrm{not}\: \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{R}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}}+\:\frac{\mathrm{1}}{\mathrm{9}}\right)^{\mathrm{2}} \:=\: \\ $$$$\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{R}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{9}^{\mathrm{2}} }\right) \\ $$$$\mathrm{Descartes}\:\mathrm{theorem} \\ $$
Commented by mr W last updated on 27/Mar/20
see Q77681
$${see}\:{Q}\mathrm{77681} \\ $$
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.
$${the}\:{big}\:{circle}\:{is}\:{touched}\:{on}\:{its}\:{inside}, \\ $$$${therefore}\:{its}\:{curvature}\:{is}\:{negative}! \\ $$$${otherwise}\:{you}\:{won}'{t}\:{be}\:{able}\:{to}\:{get}\:{a} \\ $$$${solution}. \\ $$
Commented by jagoll last updated on 27/Mar/20
ok sir. thank you
$$\mathrm{ok}\:\mathrm{sir}.\:\mathrm{thank}\:\mathrm{you} \\ $$
Commented by john santu last updated on 27/Mar/20
thank you
$${thank}\:{you} \\ $$

Leave a Reply

Your email address will not be published.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