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Question-65380

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Question Number 65380 by ajfour last updated on 29/Jul/19
Commented by ajfour last updated on 29/Jul/19
Find radius r of a circle whose center is on the circumference of the unit circle and whose arc length within the shown unit radius circle is a maximum. (what if the first condition not be imposed, but still center be outside the unit circle ?)
$${Find}\:{radius}\:{r}\:{of}\:{a}\:{circle}\:{whose} \\ $$$${center}\:{is}\:{on}\:{the}\:{circumference} \\ $$$${of}\:{the}\:{unit}\:{circle}\:{and}\:{whose} \\ $$$${arc}\:{length}\:{within}\:{the}\:{shown} \\ $$$${unit}\:{radius}\:{circle}\:{is}\:{a}\:{maximum}. \\ $$$$\left({what}\:{if}\:{the}\:{first}\:{condition}\:{not}\right. \\ $$$${be}\:{imposed},\:{but}\:{still}\:{center}\:{be} \\ $$$$\left.{outside}\:{the}\:{unit}\:{circle}\:?\right) \\ $$
Answered by ajfour last updated on 29/Jul/19
Commented by ajfour last updated on 29/Jul/19
x^2 +y^2 =1 x=rsin θ , y=−c+rcos θ ⇒ r^2 −2crcos θ+c^2 =1 cos θ=((r^2 +c^2 −1)/(2cr)) L=2rcos^(−1) (((r^2 +c^2 −1)/(2cr))) ((∂(L/2))/∂c)=0 ⇒ (r/( (√(1−(((r^2 +c^2 −1)/(2cr)))^2 ))))×(1/(2r))×(((c^2 −r^2 +1))/c^2 ) ⇒ c^2 =r^2 −1 ....(i) ((∂(L/2))/∂r)=0 ⇒ cos^(−1) (((r^2 +c^2 −1)/(2cr)))=r×(1/( (√(1−(((r^2 +c^2 −1)/(2cr)))^2 ))))×(1/(2c))×(((r^2 −c^2 +1))/r^2 ) using (i) in above eq. cos^(−1) (((√(r^2 −1))/r))=(1/( (√(r^2 −1)))) ......
$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1} \\ $$$${x}={r}\mathrm{sin}\:\theta\:,\:{y}=−{c}+{r}\mathrm{cos}\:\theta \\ $$$$\Rightarrow\:{r}^{\mathrm{2}} −\mathrm{2}{cr}\mathrm{cos}\:\theta+{c}^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{cos}\:\theta=\frac{{r}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{cr}} \\ $$$${L}=\mathrm{2}{r}\mathrm{cos}^{−\mathrm{1}} \left(\frac{{r}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{cr}}\right) \\ $$$$\frac{\partial\left({L}/\mathrm{2}\right)}{\partial{c}}=\mathrm{0}\:\:\Rightarrow \\ $$$$\frac{{r}}{\:\sqrt{\mathrm{1}−\left(\frac{{r}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{cr}}\right)^{\mathrm{2}} }}×\frac{\mathrm{1}}{\mathrm{2}{r}}×\frac{\left({c}^{\mathrm{2}} −{r}^{\mathrm{2}} +\mathrm{1}\right)}{{c}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:{c}^{\mathrm{2}} ={r}^{\mathrm{2}} −\mathrm{1}\:\:\:\:\:….\left({i}\right) \\ $$$$\frac{\partial\left({L}/\mathrm{2}\right)}{\partial{r}}=\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{cos}^{−\mathrm{1}} \left(\frac{{r}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{cr}}\right)={r}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\left(\frac{{r}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{cr}}\right)^{\mathrm{2}} }}×\frac{\mathrm{1}}{\mathrm{2}{c}}×\frac{\left({r}^{\mathrm{2}} −{c}^{\mathrm{2}} +\mathrm{1}\right)}{{r}^{\mathrm{2}} } \\ $$$${using}\:\left({i}\right)\:{in}\:{above}\:{eq}. \\ $$$$\mathrm{cos}^{−\mathrm{1}} \left(\frac{\sqrt{{r}^{\mathrm{2}} −\mathrm{1}}}{{r}}\right)=\frac{\mathrm{1}}{\:\sqrt{{r}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$…… \\ $$

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