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Question Number 71348 by ajfour last updated on 13/Oct/19

Commented by ajfour last updated on 13/Oct/19

$$Ifthecirclehasunitradiusand$$$$eachparthavesamearea,find$$$$radiusofcirculararc.$$

Commented by ajfour last updated on 14/Oct/19

Commented by ajfour last updated on 14/Oct/19

$$letcenterofsmallcirclebeorigin.$$$$p^2+q^2=1$$$$p^2+{(q+1+k)}^2=R^2$$$$\Rightarrow (1+k)(2q+1+k)=R^2-1$$$$\Rightarrow q=\frac{1}{2}[\frac{R^2-1}{1+k}-(1+k)]<0$$$$\Rightarrow p=\sqrt{1-\frac{1}{4}{[\frac{R^2-1}{1+k}-(1+k)]}^2}$$$$let\theta ={\tan }^{-1}\frac{\mid q\mid }{p}$$$$A_1=\frac{(\pi /2-\theta )}{2};$$$$let\beta ={\tan }^{-1}\frac{p}{q+1+k}$$$$A_2=\frac{R^2\beta }{2}\mathrm{\&}{A}_3=\frac{p(1+k)}{2}$$$$Now{A}_1-(A_2-A_3)=\frac{\pi }{4}\Rightarrow$$$$\frac{(\pi /2-\theta )}{2}-\frac{R^2\beta }{2}+\frac{p(1+k)}{2}=\frac{\pi }{4}$$$$A_4=\frac{R^2}{2}{\cos }^{-1}\frac{k}{R}-\frac{k\sqrt{R^2-k^2}}{2}=\frac{\pi }{2}$$$$\theta ={\tan }^{-1}\left(\frac{-\frac{1}{2}[\frac{R^2-1}{1+k}-(1+k)]}{\sqrt{1-\frac{1}{4}{[\frac{R^2-1}{1+k}-(1+k)]}^2}}\right)$$$$\beta ={\tan }^{-1}\frac{p}{q+1+k}$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}.$$

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