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Question Number 142603 by mohammad17 last updated on 02/Jun/21

Answered by mr W last updated on 03/Jun/21

Commented by mr W last updated on 03/Jun/21

$$CE=\sqrt{{(R-r)}^2-r^2}=\sqrt{R^2-2Rr}$$$$FE=R+\sqrt{R^2-2Rr}=FA=1$$$$\Rightarrow R^2-2Rr={(1-R)}^2$$$$\Rightarrow R^2-2Rr=1-2R+R^2$$$$\Rightarrow 2R(1-r)=1$$$$\Rightarrow r=1-\frac{1}{2R}$$$$$$$$\cos \theta =\frac{R}{1}=R$$$$\tan \frac{\theta }{2}=\frac{r}{1}=r$$$$\Rightarrow R=\frac{1-r^2}{1+r^2}=\frac{2}{1+r^2}-1$$$$\Rightarrow (R+1)(1+r^2)=2$$$$\Rightarrow (R+1)[1+{(1-\frac{1}{2R})}^2]=2$$$$\Rightarrow (R+1)[4R^2+{(2R-1)}^2]=8R^2$$$$\Rightarrow 8R^3-4R^2-3R+1=0$$$$\Rightarrow R\approx 0.7757$$

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