Question Number 209813 by mr W last updated on 22/Jul/24
Answered by mahdipoor last updated on 22/Jul/24
$${R}=\:{radius}\:{red}\:{circle} \\ $$$${r}\:=\:{radius}\:{white}\:{circle} \\ $$$$\begin{cases}{\mathrm{2}{R}=\mathrm{8}+\mathrm{2}+\mathrm{2}{r}}\\{{C}_{\mathrm{1}} {C}_{\mathrm{2}} =\sqrt{{r}^{\mathrm{2}} +\left({R}−\left({r}+\mathrm{2}\right)\right)^{\mathrm{2}} }={R}−{r}}\end{cases} \\ $$$$\Rightarrow{R}^{\mathrm{2}} +\mathrm{2}{r}^{\mathrm{2}} +\mathrm{4}+\mathrm{4}{r}−\mathrm{2}{Rr}−\mathrm{4}{R}={R}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{Rr} \\ $$$$\Rightarrow{r}^{\mathrm{2}} +\mathrm{4}{r}+\mathrm{4}=\mathrm{4}{R}=\mathrm{20}+\mathrm{4}{r}\:\Rightarrow{r}^{\mathrm{2}} =\mathrm{16} \\ $$$$\Rightarrow{r}=\mathrm{4}\Rightarrow{R}=\mathrm{9}\Rightarrow{S}_{{red}} =\frac{\pi}{\mathrm{2}}{R}^{\mathrm{2}} −\pi{r}^{\mathrm{2}} =\frac{\mathrm{49}\pi}{\mathrm{2}}\: \\ $$$${thank}\:{mr}.{w}\:,\:{i}\:{edited}. \\ $$
Commented by mr W last updated on 22/Jul/24
$${great}! \\ $$
Answered by cherokeesay last updated on 22/Jul/24
Commented by mr W last updated on 22/Jul/24
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