Question-86063

Question Number 86063 by john santu last updated on 26/Mar/20

Answered by mr W last updated on 26/Mar/20

{(\frac{1}{R} + \frac{1}{3} + \frac{1}{6} - \frac{1}{9})}^{2} = 2 (\frac{1}{R^{2}} + \frac{1}{3^{2}} + \frac{1}{6^{2}} + \frac{1}{9^{2}})

\Rightarrow R = \frac{18}{7}

Commented by jagoll last updated on 26/Mar/20

how sir, got it ?

explain me . please

Commented by jagoll last updated on 27/Mar/20

sir it not

{(\frac{1}{R} + \frac{1}{3} + \frac{1}{6} + \frac{1}{9})}^{2} =

2 (\frac{1}{R^{2}} + \frac{1}{3^{2}} + \frac{1}{6^{2}} + \frac{1}{9^{2}})

Descartes theorem

Commented by mr W last updated on 27/Mar/20

s e e Q 77681

Commented by mr W last updated on 27/Mar/20

t h e b i g c i r c l e i s t o u c h e d o n i t s i n s i d e,

t h e r e f o r e i t s c u r v a t u r e i s n e g a t i v e!

o t h e r w i s e y o u {w o n}^{'} t b e a b l e t o g e t a

s o l u t i o n .

Commented by jagoll last updated on 27/Mar/20

ok sir . thank you

Commented by john santu last updated on 27/Mar/20

t h a n k y o u