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Question-197916

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Question Number 197916 by sonukgindia last updated on 04/Oct/23
Answered by MM42 last updated on 04/Oct/23
Commented by MM42 last updated on 05/Oct/23
let OB=r & CD=x 9=(2+2r+x)^2 +1⇒2r+x=(√8)−2⇒2r=(√8)−2−x (1+r)^2 =(x+r)^2 +1⇒2r=2rx+x^2 (√8)−2−x=(√8)x−2x−x^2 +x^2 ((√8)−1)x=(√8)−2⇒x=(((√8)−2)/( (√8)−1)) ⇒r=((12−4(√8))/( (√8)−1)) ✓
$${let}\:\:{OB}={r}\:\:\&\:{CD}={x} \\ $$$$\mathrm{9}=\left(\mathrm{2}+\mathrm{2}{r}+{x}\right)^{\mathrm{2}} +\mathrm{1}\Rightarrow\mathrm{2}{r}+{x}=\sqrt{\mathrm{8}}−\mathrm{2}\Rightarrow\mathrm{2}{r}=\sqrt{\mathrm{8}}−\mathrm{2}−{x} \\ $$$$\left(\mathrm{1}+{r}\right)^{\mathrm{2}} =\left({x}+{r}\right)^{\mathrm{2}} +\mathrm{1}\Rightarrow\mathrm{2}{r}=\mathrm{2}{rx}+{x}^{\mathrm{2}} \\ $$$$\sqrt{\mathrm{8}}−\mathrm{2}−{x}=\sqrt{\mathrm{8}}{x}−\mathrm{2}{x}−{x}^{\mathrm{2}} +{x}^{\mathrm{2}} \\ $$$$\left(\sqrt{\mathrm{8}}−\mathrm{1}\right){x}=\sqrt{\mathrm{8}}−\mathrm{2}\Rightarrow{x}=\frac{\sqrt{\mathrm{8}}−\mathrm{2}}{\:\sqrt{\mathrm{8}}−\mathrm{1}} \\ $$$$\Rightarrow{r}=\frac{\mathrm{12}−\mathrm{4}\sqrt{\mathrm{8}}}{\:\sqrt{\mathrm{8}}−\mathrm{1}}\:\:\checkmark \\ $$
Answered by mr W last updated on 05/Oct/23
generally ((1/R)+(1/r_1 )+(1/r_2 )+(1/r_3 ))^2 =2((1/R^2 )+(1/r_1 ^2 )+(1/r_2 ^2 )+(1/r_3 ^2 )) ((1/R)+(1/1)+(1/1)+(1/2))^2 =2((1/R^2 )+(1/1^2 )+(1/1^2 )+(1/2^2 )) (1/R^2 )−(5/R)−(7/4)=0 (1/R)=((5±4(√2))/2) ⇒R=(2/(5±4(√2)))=((8(√2)−10)/7), −((8(√2)+10)/7) radius of small circle =((8(√2)−10)/7)≈0.188 radius of big circle =((8(√2)+10)/7)≈3.044
$${generally} \\ $$$$\left(\frac{\mathrm{1}}{{R}}+\frac{\mathrm{1}}{{r}_{\mathrm{1}} }+\frac{\mathrm{1}}{{r}_{\mathrm{2}} }+\frac{\mathrm{1}}{{r}_{\mathrm{3}} }\right)^{\mathrm{2}} =\mathrm{2}\left(\frac{\mathrm{1}}{{R}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}_{\mathrm{1}} ^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}_{\mathrm{2}} ^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}_{\mathrm{3}} ^{\mathrm{2}} }\right) \\ $$$$\left(\frac{\mathrm{1}}{{R}}+\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{2}\left(\frac{\mathrm{1}}{{R}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\right) \\ $$$$\frac{\mathrm{1}}{{R}^{\mathrm{2}} }−\frac{\mathrm{5}}{{R}}−\frac{\mathrm{7}}{\mathrm{4}}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{{R}}=\frac{\mathrm{5}\pm\mathrm{4}\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\Rightarrow{R}=\frac{\mathrm{2}}{\mathrm{5}\pm\mathrm{4}\sqrt{\mathrm{2}}}=\frac{\mathrm{8}\sqrt{\mathrm{2}}−\mathrm{10}}{\mathrm{7}},\:−\frac{\mathrm{8}\sqrt{\mathrm{2}}+\mathrm{10}}{\mathrm{7}} \\ $$$${radius}\:{of}\:{small}\:{circle}\:=\frac{\mathrm{8}\sqrt{\mathrm{2}}−\mathrm{10}}{\mathrm{7}}\approx\mathrm{0}.\mathrm{188} \\ $$$${radius}\:{of}\:{big}\:{circle}\:=\frac{\mathrm{8}\sqrt{\mathrm{2}}+\mathrm{10}}{\mathrm{7}}\approx\mathrm{3}.\mathrm{044} \\ $$

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