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Question-197238

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Question Number 197238 by sonukgindia last updated on 10/Sep/23
Commented by Frix last updated on 12/Sep/23
No useful exact solution. Set the radius of the quarter circle =1 ⇒ Radius of the semi circle ≈.717910784 r≈.205269551 R≈.342040196 (R/r)≈1.66629777 Let p the x−value of the center of the right small circle, q the y−value of the center of the left one and o the x−value of the center of the circle in the mid p is the positive solution of p^4 +2p^3 +p^2 −4p−2=0 p≈1.18766119 q=(√(2−p^2 )) o=−((p(p−(√((p−1)(3p+1))))/(2p^2 −2p−1)) Radius of semi circle =−(p^2 /(p^2 −2p−1)) r=((p^2 −1)/2) R=((−(p−1)(2p+1)(3p+1)+2p^3 (√((p−1)(3p+1))))/(2(2p^2 −2p−1)^2 ))
$$\mathrm{No}\:\mathrm{useful}\:\mathrm{exact}\:\mathrm{solution}. \\ $$$$\mathrm{Set}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{quarter}\:\mathrm{circle}\:=\mathrm{1} \\ $$$$\Rightarrow \\ $$$$\mathrm{Radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{semi}\:\mathrm{circle}\:\approx.\mathrm{717910784} \\ $$$${r}\approx.\mathrm{205269551} \\ $$$${R}\approx.\mathrm{342040196} \\ $$$$\frac{{R}}{{r}}\approx\mathrm{1}.\mathrm{66629777} \\ $$$$ \\ $$$$\mathrm{Let}\:{p}\:\mathrm{the}\:{x}−\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{center}\:\mathrm{of}\:\mathrm{the}\:\mathrm{right} \\ $$$$\mathrm{small}\:\mathrm{circle},\:{q}\:\mathrm{the}\:{y}−\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{center}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{left}\:\mathrm{one}\:\mathrm{and}\:{o}\:\mathrm{the}\:{x}−\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{center} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{in}\:\mathrm{the}\:\mathrm{mid} \\ $$$${p}\:\mathrm{is}\:\mathrm{the}\:\mathrm{positive}\:\mathrm{solution}\:\mathrm{of} \\ $$$${p}^{\mathrm{4}} +\mathrm{2}{p}^{\mathrm{3}} +{p}^{\mathrm{2}} −\mathrm{4}{p}−\mathrm{2}=\mathrm{0} \\ $$$${p}\approx\mathrm{1}.\mathrm{18766119} \\ $$$${q}=\sqrt{\mathrm{2}−{p}^{\mathrm{2}} } \\ $$$${o}=−\frac{{p}\left({p}−\sqrt{\left({p}−\mathrm{1}\right)\left(\mathrm{3}{p}+\mathrm{1}\right)}\right.}{\mathrm{2}{p}^{\mathrm{2}} −\mathrm{2}{p}−\mathrm{1}} \\ $$$$\mathrm{Radius}\:\mathrm{of}\:\mathrm{semi}\:\mathrm{circle}\:=−\frac{{p}^{\mathrm{2}} }{{p}^{\mathrm{2}} −\mathrm{2}{p}−\mathrm{1}} \\ $$$${r}=\frac{{p}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}} \\ $$$${R}=\frac{−\left({p}−\mathrm{1}\right)\left(\mathrm{2}{p}+\mathrm{1}\right)\left(\mathrm{3}{p}+\mathrm{1}\right)+\mathrm{2}{p}^{\mathrm{3}} \sqrt{\left({p}−\mathrm{1}\right)\left(\mathrm{3}{p}+\mathrm{1}\right)}}{\mathrm{2}\left(\mathrm{2}{p}^{\mathrm{2}} −\mathrm{2}{p}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$

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