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

Commented by ajfour last updated on 19/Aug/20

$$Findb/a.$$

Answered by ajfour last updated on 19/Aug/20

Commented by ajfour last updated on 19/Aug/20

$$LetEq.ofleftsemicircleis$$$${(x+1)}^2+{(y+1)}^2=4$$$$Itsintersectionwithy=xline$$$$gives{(\frac{a}{\sqrt{2}}+1)}^2=2$$$$\Rightarrow a=\sqrt{2}(\sqrt{2}-1)=2-\sqrt{2}$$$$letcenterofrightbluecirclebe$$$$C(h,k)\mathrm{\&}k=1-b$$$$\Rightarrow {(h+1)}^2+{(k+1)}^2={(b+2)}^2\mathrm{\&}$$$${(h-1)}^2+{(1-k)}^2={(2-b)}^2$$$$\Rightarrow {(h+1)}^2=8b$$$${(h-1)}^2=4-4b$$$$\Rightarrow 4h=12b-4\Rightarrow h=3b-1$$$$\Rightarrow 9b^2=8b\Rightarrow b=\frac{8}{9}$$$$hence\frac{b}{a}=\frac{8}{9(2-\sqrt{2})}=\frac{4}{9}(2+\sqrt{2}).$$

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

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

$$R=radiusofsemicircles$$$$a=R-\frac{R}{\sqrt{2}}=\frac{(2-\sqrt{2})R}{2}$$$$AB=\sqrt{{(R-b)}^2-b^2}=\sqrt{R^2-2Rb}$$$${AC}^2+{(EC-DA)}^2={DE}^2$$$${(\sqrt{R^2-2Rb}+R)}^2+{(R-b)}^2={(R+b)}^2$$$$\sqrt{R^2-2Rb}=3b-R$$$$9b-4R=0$$$$\Rightarrow b=\frac{4R}{9}$$$$\frac{b}{a}=\frac{4\times 2}{9\times (2-\sqrt{2})}=\frac{4(2+\sqrt{2})}{9}\approx 1.517$$

Commented by ajfour last updated on 19/Aug/20

$$UnderstoodSir,yourwayis$$$$simpler,thanksalot!$$

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