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Question Number 28399 by ajfour last updated on 25/Jan/18

Answered by ajfour last updated on 25/Jan/18

$$eq.ofcircle:$$$$x=r+r\cos \theta ,y=r+r\sin \theta$$$$eq.ofline:$$$$\frac{x}{a}+\frac{y}{b}=1$$$$intersectionpoints:{\theta }_1,{\theta }_{2}obeys$$$$\frac{1+\cos \theta }{a}+\frac{1+\sin \theta }{b}=\frac{1}{r}$$$$\frac{\cos \theta }{a}+\frac{\sin \theta }{b}=\frac{1}{r}-\frac{a+b}{ab}$$$$let△\theta ={\theta }_2-{\theta }_1$$$$\Rightarrow b\cos \frac{△\theta }{2}\cos \left(\frac{{\theta }_1+{\theta }_2}{2}\right)+$$$$a\sin \left(\frac{{\theta }_1+{\theta }_2}{2}\right)\cos \frac{△\theta }{2}=\frac{ab}{r}-(a+b)$$$$Since\tan \left(\frac{{\theta }_1+{\theta }_2}{2}\right)=\frac{a}{b}$$$$\frac{b^2}{\sqrt{a^2+b^2}}\cos \frac{△\theta }{2}+\frac{a^2}{\sqrt{a^2+b^2}}\cos \frac{△\theta }{2}=\frac{ab}{r}-(a+b)$$$$\sqrt{a^2+b^2}\cos \left(\frac{△\theta }{2}\right)=\frac{ab}{r}-(a+b)$$$$\frac{△\theta }{2}={\cos }^{-1}\left[\frac{1}{\sqrt{a^2+b^2}}(\frac{ab}{r}-a-b)\right]$$$$RequiredArea=$$$$\frac{r^2△\theta }{2}-\frac{1}{2}\times \left(2r\sin \frac{△\theta }{2}\right)\left(r\cos \frac{△\theta }{2}\right)$$$$orA=\frac{r^2△\theta }{2}(1-\frac{\sin △\theta }{△\theta }).$$

Commented by mrW2 last updated on 25/Jan/18

$$Iwouldtrylikethis:$$$$\frac{\cos \theta }{a}+\frac{\sin \theta }{b}=\frac{1}{r}-\frac{a+b}{ab}$$$$\Rightarrow a\sin \theta +b\cos \theta =\frac{ab}{r}-a-b$$$$\frac{a}{\sqrt{a^2+b^2}}\sin \theta +\frac{b}{\sqrt{a^2+b^2}}\cos \theta =\frac{1}{\sqrt{a^2+b^2}}(\frac{ab}{r}-a-b)$$$$\sin \alpha \sin \theta +\cos \alpha \cos \theta =\frac{1}{\sqrt{a^2+b^2}}(\frac{ab}{r}-a-b)$$$$with\alpha ={\tan }^{-1}\frac{a}{b}$$$$\Rightarrow \cos (\theta -\alpha )=\frac{1}{\sqrt{a^2+b^2}}(\frac{ab}{r}-a-b)$$$$\theta ={\tan }^{-1}\frac{a}{b}\pm {\cos }^{-1}\left[\frac{1}{\sqrt{a^2+b^2}}(\frac{ab}{r}-a-b)\right]$$$$\mathrm{\Delta }\theta ={\theta }_2-{\theta }_1=2{\cos }^{-1}\left[\frac{1}{\sqrt{a^2+b^2}}(\frac{ab}{r}-a-b)\right]$$$$......$$$$A=\frac{r^2}{2}(\mathrm{\Delta }\theta -\sin \mathrm{\Delta }\theta )$$

Commented by ajfour last updated on 25/Jan/18

$$thanksverymuchSir,icorrected.$$

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