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Question Number 40489 by ajfour last updated on 22/Jul/18

Commented by ajfour last updated on 22/Jul/18

$$IntermsofradiusofcircleR,$$$$finda,andb.$$

Commented by behi83417@gmail.com last updated on 23/Jul/18

$$BC.AF=AB.CD$$$$cosOAD=\frac{a}{b},OD=\sqrt{b^2-a^2}$$$$tgOED=\frac{\sqrt{b^2-a^2}}{OE}\Rightarrow OE=\sqrt{b^2-a^2}.\frac{\sqrt{b^2-a^2}}{a}=\frac{b^2-a^2}{a}$$$$AE=a+\frac{b^2-a^2}{a}=\frac{b^2}{a},{DE}^2=b^2-a^2+\frac{{(b^2-a^2)}^2}{a^2}=$$$$=\frac{(b^2-a^2)(a^2+b^2-a^2)}{a^2}\Rightarrow DE=\frac{b}{a}\sqrt{b^2-a^2}$$$$\frac{DE}{EC}=\frac{OE}{EF}\Rightarrow EF=\frac{b^2-a^2}{a}.\frac{b}{\frac{b}{a}\sqrt{b^2-a^2}}=\sqrt{b^2-a^2}$$$$\frac{a}{b}=\frac{b+BD}{a+\frac{b^2-a^2}{a}+\sqrt{b^2-a^2}}\Rightarrow BD=\frac{b^2+a\sqrt{b^2-a^2}}{b}-b$$$$=\frac{a}{b}\sqrt{b^2-a^2}$$$$cosBDO=cos(90+EDO)=-sinEDO=$$$$=-\frac{OE}{ED}=-\frac{\frac{b^2-a^2}{a}}{\frac{b}{a}\sqrt{b^2-a^2}}=-\frac{\sqrt{b^2-a^2}}{b}$$$$cosBDO=-\frac{\sqrt{b^2-a^2}}{b}$$$$R^2={OD}^2+{BD}^2-2OD.BD.cosBDO=$$$$=(b^2-a^2)+\frac{a^2}{b^2}(b^2-a^2)+2\sqrt{b^2-a^2}.\frac{a}{b}\sqrt{b^2-a^2}.\frac{\sqrt{b^2-a^2}}{b}=$$$$=(b^2-a^2)(1+\frac{a^2}{b^2}+2\frac{a}{b^2}\sqrt{b^2-a^2})=$$$$=\frac{b^2-a^2}{b^2}(b^2+a^2+2a\sqrt{b^2-a^2})$$$$\Rightarrow R=\frac{\sqrt{b^2+a^2+2a\sqrt{b^2-a^2}}}{b}.\sqrt{b^2-a^2}.$$$$BC.\frac{b^2+a\sqrt{b^2-a^2}}{a}=\frac{b^2+a\sqrt{b^2-a^2}}{b}.(b+\frac{b}{a}\sqrt{b^2-a^2})$$$$BC=a+\sqrt{b^2-a^2}.$$$$OF.BC=R^2.sinCOB\Rightarrow$$$$sinCOB=\frac{(a+\sqrt{b^2-a^2})(\frac{b^2-a^2}{a}+\sqrt{b^2-a^2})}{R^2}=$$$$=\frac{{(a+\sqrt{b^2-a^2})}^2.\sqrt{b^2-a^2}}{{aR}^2}$$

Commented by ajfour last updated on 23/Jul/18

$$ThanksSir.$$

Answered by ajfour last updated on 23/Jul/18

$$let\mathrm{\angle }OAB=\mathrm{\angle }ECF=\theta$$$$b\cos \theta =a....\left(i\right)$$$$AE=b\sec \theta$$$$AlsoCF=BF=b\cos \theta =a$$$$EF=b\sin \theta =AF-AE$$$$\Rightarrow b\sin \theta =a\cot \theta -b\sec \theta ..\left(ii\right)$$$$OF=\sqrt{R^2-{BF}^2}=\sqrt{R^2-a^2}$$$$OF=OE+EF$$$$\Rightarrow \sqrt{R^2-a^2}=b\sin {}^2\theta +b\sin \theta ..\left(iii\right)$$$$using\left(i\right)in\left(ii\right):$$$$a\sec \theta \sin \theta =a\cot \theta -a\sec {}^2\theta$$$$\Rightarrow \tan \theta =\cot \theta -1-\tan {}^2\theta$$$$let\tan \theta =m;then$$$$m^2=1-m-m^3$$$$or{m}^3+m^2+m=1$$$$\Rightarrow m\approx 0.54369$$$$using\left(i\right)in\left(iii\right)$$$$\sqrt{R^2-a^2}=a\sec \theta \sin {}^2\theta +a\sec \theta \sin \theta$$$$=a\tan \theta (1+\sin \theta )$$$$R^2-a^2=a^2{m}^2{(1+\frac{m}{\sqrt{1+m^2}})}^2$$$$\Rightarrow a^2=\frac{R^2}{1+m^2{(1+\frac{m}{\sqrt{1+m^2}})}^2}$$$$\Rightarrow a\approx 0.7743R$$$$b=a\sqrt{1+m^2}\approx 0.88R.$$

Commented by MJS last updated on 23/Jul/18

$$\mathrm{I}\mathrm{get}\mathrm{the}\mathrm{same}\mathrm{value}\mathrm{for}\theta$$$$\mathrm{so}\mathrm{either}\mathrm{me}\mathrm{or}\mathrm{you}\mathrm{made}\mathrm{a}\mathrm{mistake}\mathrm{somewhere}$$

Answered by MJS last updated on 23/Jul/18

$$R=1$$$$\mathrm{\angle }DAO=\mathrm{\angle }ECF\Rightarrow \mid CF\mid =\mid BF\mid =a$$$$\mid OF\mid =\sqrt{1-a^2}$$$$\mid OD\mid =\mid BF\mid \frac{\mid AO\mid }{\mid AF\mid }=\frac{a^2}{a+\sqrt{1-a^2}}$$$$\stackrel{\rightharpoonup }{AB}=B-A=\left(\begin{array}{c}a+\sqrt{1-a^2}\\ a\end{array}\right)$$$$\stackrel{\rightharpoonup }{CD}=D-C=\left(\begin{array}{c}-\sqrt{1-a^2}\\ a\frac{2a+\sqrt{1-a^2}}{a+\sqrt{1-a^2}}\end{array}\right)$$$$\stackrel{\rightharpoonup }{AB}\mathrm{\perp }\stackrel{\rightharpoonup }{CD}$$$$(a+\sqrt{1-a^2})(-\sqrt{1-a^2})+a^2\frac{2a+\sqrt{1-a^2}}{a+\sqrt{1-a^2}}=0$$$$a^6-\frac{19}{17}{a}^4+\frac{7}{17}{a}^2-\frac{1}{17}=0$$$$a=\sqrt{x}$$$$x^3-\frac{19}{17}{x}^2+\frac{7}{17}x-\frac{1}{17}=0$$$$x=z+\frac{19}{51}$$$$z^3-\frac{4}{867}z-\frac{1172}{132651}=0$$$$z\approx .214167$$$$x\approx .586716$$$$a\approx .765974$$$$\mid OD\mid \approx .416452$$$$b\approx .871865$$$$$$$$a\approx .765974R$$$$b\approx .871865R$$

Commented by ajfour last updated on 23/Jul/18

$$matchesclosetomyanswer;$$$$thankyouSir.$$

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