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Question Number 58547 by naka3546 last updated on 24/Apr/19

Commented by naka3546 last updated on 24/Apr/19

find the radius of purple circle in r .

$${find}\:\:{the}\:\:{radius}\:\:{of}\:\:{purple}\:\:{circle}\:\:{in}\:\:{r}\:. \\ $$

Answered by mr W last updated on 24/Apr/19

Commented by mr W last updated on 24/Apr/19

OA=(√((1+r)^2 −r^2 ))−1=(√(1+2r))−1 OB=2−r OB^2 =OA^2 +r^2 (2−r)^2 =((√(1+2r))−1)^2 +r^2 4−4r+r^2 =1+2r−2(√(1+2r))+1+r^2 3r−1=(√(1+2r)) 9r^2 −6r+1=1+2r ⇒r=(8/9)

$${OA}=\sqrt{\left(\mathrm{1}+{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} }−\mathrm{1}=\sqrt{\mathrm{1}+\mathrm{2}{r}}−\mathrm{1} \\ $$$${OB}=\mathrm{2}−{r} \\ $$$${OB}^{\mathrm{2}} ={OA}^{\mathrm{2}} +{r}^{\mathrm{2}} \\ $$$$\left(\mathrm{2}−{r}\right)^{\mathrm{2}} =\left(\sqrt{\mathrm{1}+\mathrm{2}{r}}−\mathrm{1}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} \\ $$$$\mathrm{4}−\mathrm{4}{r}+{r}^{\mathrm{2}} =\mathrm{1}+\mathrm{2}{r}−\mathrm{2}\sqrt{\mathrm{1}+\mathrm{2}{r}}+\mathrm{1}+{r}^{\mathrm{2}} \\ $$$$\mathrm{3}{r}−\mathrm{1}=\sqrt{\mathrm{1}+\mathrm{2}{r}} \\ $$$$\mathrm{9}{r}^{\mathrm{2}} −\mathrm{6}{r}+\mathrm{1}=\mathrm{1}+\mathrm{2}{r} \\ $$$$\Rightarrow{r}=\frac{\mathrm{8}}{\mathrm{9}} \\ $$

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