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Question Number 108868 by ajfour last updated on 19/Aug/20

Commented by ajfour last updated on 19/Aug/20

Find b/a.

$${Find}\:{b}/{a}. \\ $$

Answered by ajfour last updated on 19/Aug/20

Commented by ajfour last updated on 19/Aug/20

Let Eq. of left semicircle is (x+1)^2 +(y+1)^2 =4 Its intersection with y=x line gives ((a/( (√2)))+1)^2 =2 ⇒ a=(√2)((√2)−1) = 2−(√2) let center of right blue circle be C(h, k) & k=1−b ⇒ (h+1)^2 +(k+1)^2 =(b+2)^2 & (h−1)^2 +(1−k)^2 =(2−b)^2 ⇒ (h+1)^2 = 8b (h−1)^2 =4−4b ⇒ 4h=12b−4 ⇒ h=3b−1 ⇒ 9b^2 =8b ⇒ b=(8/9) hence (b/a)=(8/(9(2−(√2)))) = (4/9)(2+(√2)) .

$${Let}\:{Eq}.\:{of}\:{left}\:{semicircle}\:{is} \\ $$$$\:\:\:\:\:\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\left({y}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{4} \\ $$$${Its}\:{intersection}\:{with}\:\:{y}={x}\:{line} \\ $$$${gives}\:\:\:\left(\frac{{a}}{\:\sqrt{\mathrm{2}}}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{2} \\ $$$$\Rightarrow\:\:\:\:\:{a}=\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\:=\:\mathrm{2}−\sqrt{\mathrm{2}} \\ $$$${let}\:{center}\:{of}\:{right}\:{blue}\:{circle}\:{be} \\ $$$${C}\left({h},\:{k}\right)\:\:\:\:\&\:\:\:{k}=\mathrm{1}−{b} \\ $$$$\Rightarrow\:\:\:\left({h}+\mathrm{1}\right)^{\mathrm{2}} +\left({k}+\mathrm{1}\right)^{\mathrm{2}} =\left({b}+\mathrm{2}\right)^{\mathrm{2}} \:\:\:\:\& \\ $$$$\:\:\:\:\:\:\:\:\:\left({h}−\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{1}−{k}\right)^{\mathrm{2}} =\left(\mathrm{2}−{b}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:\left({h}+\mathrm{1}\right)^{\mathrm{2}} =\:\mathrm{8}{b} \\ $$$$\:\:\:\:\:\:\:\:\:\left({h}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{4}−\mathrm{4}{b} \\ $$$$\Rightarrow\:\:\:\mathrm{4}{h}=\mathrm{12}{b}−\mathrm{4}\:\:\:\Rightarrow\:\:{h}=\mathrm{3}{b}−\mathrm{1} \\ $$$$\Rightarrow\:\:\:\:\:\mathrm{9}{b}^{\mathrm{2}} =\mathrm{8}{b}\:\:\:\:\Rightarrow\:\:\:{b}=\frac{\mathrm{8}}{\mathrm{9}} \\ $$$$\:{hence}\:\:\frac{{b}}{{a}}=\frac{\mathrm{8}}{\mathrm{9}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)}\:=\:\frac{\mathrm{4}}{\mathrm{9}}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)\:. \\ $$

Answered by mr W last updated on 19/Aug/20

Commented by mr W last updated on 19/Aug/20

R=radius of semicircles a=R−(R/( (√2)))=(((2−(√2))R)/2) AB=(√((R−b)^2 −b^2 ))=(√(R^2 −2Rb)) AC^2 +(EC−DA)^2 =DE^2 ((√(R^2 −2Rb))+R)^2 +(R−b)^2 =(R+b)^2 (√(R^2 −2Rb))=3b−R 9b−4R=0 ⇒b=((4R)/9) (b/a)=((4×2)/(9×(2−(√2))))=((4(2+(√2)))/9)≈1.517

$${R}={radius}\:{of}\:{semicircles} \\ $$$${a}={R}−\frac{{R}}{\:\sqrt{\mathrm{2}}}=\frac{\left(\mathrm{2}−\sqrt{\mathrm{2}}\right){R}}{\mathrm{2}} \\ $$$${AB}=\sqrt{\left({R}−{b}\right)^{\mathrm{2}} −{b}^{\mathrm{2}} }=\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Rb}} \\ $$$${AC}^{\mathrm{2}} +\left({EC}−{DA}\right)^{\mathrm{2}} ={DE}^{\mathrm{2}} \\ $$$$\left(\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Rb}}+{R}\right)^{\mathrm{2}} +\left({R}−{b}\right)^{\mathrm{2}} =\left({R}+{b}\right)^{\mathrm{2}} \\ $$$$\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Rb}}=\mathrm{3}{b}−{R} \\ $$$$\mathrm{9}{b}−\mathrm{4}{R}=\mathrm{0} \\ $$$$\Rightarrow{b}=\frac{\mathrm{4}{R}}{\mathrm{9}} \\ $$$$\frac{{b}}{{a}}=\frac{\mathrm{4}×\mathrm{2}}{\mathrm{9}×\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)}=\frac{\mathrm{4}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)}{\mathrm{9}}\approx\mathrm{1}.\mathrm{517} \\ $$

Commented by ajfour last updated on 19/Aug/20

Understood Sir, your way is simpler, thanks a lot!

$${Understood}\:{Sir},\:{your}\:{way}\:{is} \\ $$$${simpler},\:{thanks}\:{a}\:{lot}! \\ $$

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