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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

$$\mathrm{how}\mathrm{sir},\mathrm{got}\mathrm{it}?$$$$\mathrm{explain}\mathrm{me}.\mathrm{please}$$$$$$

Commented by jagoll last updated on 27/Mar/20

$$\mathrm{sir}\mathrm{it}\mathrm{not}$$$${(\frac{1}{\mathrm{R}}+\frac{1}{3}+\frac{1}{6}+\frac{1}{9})}^2=$$$$2(\frac{1}{{\mathrm{R}}^2}+\frac{1}{3^2}+\frac{1}{6^2}+\frac{1}{9^2})$$$$\mathrm{Descartes}\mathrm{theorem}$$

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

$$seeQ77681$$

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

$$thebigcircleistouchedonitsinside,$$$$thereforeitscurvatureisnegative!$$$$otherwiseyou{won}^{\prime }tbeabletogeta$$$$solution.$$

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

$$thankyou$$

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