Question Number 71348 by ajfour last updated on 13/Oct/19
Commented by ajfour last updated on 13/Oct/19
$${If}\:{the}\:{circle}\:{has}\:{unit}\:{radius}\:{and} \\ $$$${each}\:{part}\:{have}\:{same}\:{area},\:{find} \\ $$$${radius}\:{of}\:{circular}\:{arc}. \\ $$
Commented by ajfour last updated on 14/Oct/19
Commented by ajfour last updated on 14/Oct/19
$${let}\:{center}\:{of}\:{small}\:{circle}\:{be}\:{origin}. \\ $$$${p}^{\mathrm{2}} +{q}^{\mathrm{2}} =\mathrm{1} \\ $$$${p}^{\mathrm{2}} +\left({q}+\mathrm{1}+{k}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\Rightarrow\:\left(\mathrm{1}+{k}\right)\left(\mathrm{2}{q}+\mathrm{1}+{k}\right)={R}^{\mathrm{2}} −\mathrm{1} \\ $$$$\Rightarrow\:{q}=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{{R}^{\mathrm{2}} −\mathrm{1}}{\mathrm{1}+{k}}−\left(\mathrm{1}+{k}\right)\right]\:\:\:<\:\mathrm{0} \\ $$$$\Rightarrow\:\:{p}=\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\left[\frac{{R}^{\mathrm{2}} −\mathrm{1}}{\mathrm{1}+{k}}−\left(\mathrm{1}+{k}\right)\right]^{\mathrm{2}} } \\ $$$${let}\:\:\:\theta=\mathrm{tan}^{−\mathrm{1}} \frac{\mid{q}\mid}{{p}} \\ $$$$\:\:{A}_{\mathrm{1}} =\frac{\left(\pi/\mathrm{2}−\theta\right)}{\mathrm{2}}\:\:;\: \\ $$$${let}\:\:\beta=\mathrm{tan}^{−\mathrm{1}} \frac{{p}}{{q}+\mathrm{1}+{k}} \\ $$$${A}_{\mathrm{2}} =\frac{{R}^{\mathrm{2}} \beta}{\mathrm{2}}\:\:\&\:\:\:{A}_{\mathrm{3}} =\frac{{p}\left(\mathrm{1}+{k}\right)}{\mathrm{2}} \\ $$$${Now}\:\:\:{A}_{\mathrm{1}} −\left({A}_{\mathrm{2}} −{A}_{\mathrm{3}} \right)=\frac{\pi}{\mathrm{4}}\:\:\Rightarrow \\ $$$$\:\frac{\left(\pi/\mathrm{2}−\theta\right)}{\mathrm{2}}−\frac{{R}^{\mathrm{2}} \beta}{\mathrm{2}}+\frac{{p}\left(\mathrm{1}+{k}\right)}{\mathrm{2}}=\frac{\pi}{\mathrm{4}} \\ $$$$\:{A}_{\mathrm{4}} =\frac{{R}^{\mathrm{2}} }{\mathrm{2}}\mathrm{cos}^{−\mathrm{1}} \frac{{k}}{{R}}−\frac{{k}\sqrt{{R}^{\mathrm{2}} −{k}^{\mathrm{2}} }}{\mathrm{2}}=\frac{\pi}{\mathrm{2}} \\ $$$$\:\:\theta=\mathrm{tan}^{−\mathrm{1}} \left(\frac{−\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{{R}^{\mathrm{2}} −\mathrm{1}}{\mathrm{1}+{k}}−\left(\mathrm{1}+{k}\right)\right]}{\:\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\left[\frac{{R}^{\mathrm{2}} −\mathrm{1}}{\mathrm{1}+{k}}−\left(\mathrm{1}+{k}\right)\right]^{\mathrm{2}} }}\right) \\ $$$$\:\:\beta=\mathrm{tan}^{−\mathrm{1}} \frac{{p}}{{q}+\mathrm{1}+{k}} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_. \\ $$