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Question Number 65380 by ajfour last updated on 29/Jul/19

Commented by ajfour last updated on 29/Jul/19

$$Findradiusrofacirclewhose$$$$centerisonthecircumference$$$$oftheunitcircleandwhose$$$$arclengthwithintheshown$$$$unitradiuscircleisamaximum.$$$$(whatifthefirstconditionnot$$$$beimposed,butstillcenterbe$$$$outsidetheunitcircle?)$$

Answered by ajfour last updated on 29/Jul/19

Commented by ajfour last updated on 29/Jul/19

$$x^2+y^2=1$$$$x=r\sin \theta ,y=-c+r\cos \theta$$$$\Rightarrow r^2-2cr\cos \theta +c^2=1$$$$\cos \theta =\frac{r^2+c^2-1}{2cr}$$$$L=2r{\cos }^{-1}\left(\frac{r^2+c^2-1}{2cr}\right)$$$$\frac{\partial \left(L/2\right)}{\partial c}=0\Rightarrow$$$$\frac{r}{\sqrt{1-{\left(\frac{r^2+c^2-1}{2cr}\right)}^2}}\times \frac{1}{2r}\times \frac{(c^2-r^2+1)}{c^2}$$$$\Rightarrow c^2=r^2-1....\left(i\right)$$$$\frac{\partial \left(L/2\right)}{\partial r}=0\Rightarrow$$$${\cos }^{-1}\left(\frac{r^2+c^2-1}{2cr}\right)=r\times \frac{1}{\sqrt{1-{\left(\frac{r^2+c^2-1}{2cr}\right)}^2}}\times \frac{1}{2c}\times \frac{(r^2-c^2+1)}{r^2}$$$$using\left(i\right)inaboveeq.$$$${\cos }^{-1}\left(\frac{\sqrt{r^2-1}}{r}\right)=\frac{1}{\sqrt{r^2-1}}$$$$......$$

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