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Question Number 50460 by ajfour last updated on 16/Dec/18

Commented by ajfour last updated on 16/Dec/18

$$Find\mathit{r}intermsof\mathit{R}.$$

Commented by MJS last updated on 18/Dec/18

$$\mathrm{I}\mathrm{solved}\mathrm{this}\mathrm{for}2\mathrm{small}\mathrm{circles}\mathrm{with}r=1\mathrm{and}$$$$.\mathrm{a}\mathrm{parabola}y={ax}^2$$$$.\mathrm{a}\mathrm{circle}{x}^2+{(y-r)}^2=r^2$$$$.\mathrm{an}\mathrm{ellipse}\frac{x^2}{a^2}+\frac{{(y-b)}^2}{b^2}=1$$$$[\mathrm{this}\mathrm{one}\mathrm{is}\mathrm{tricky},\mathrm{will}\mathrm{post}\mathrm{soon}]$$$$\mathrm{will}\mathrm{also}\mathrm{try}\mathrm{for}\mathrm{a}\mathrm{hyperbola}$$

Answered by mr W last updated on 16/Dec/18

$$\frac{r}{R-r}=\frac{\sqrt{{(R+r)}^2-{(R-r)}^2}}{R+r}=\frac{2\sqrt{Rr}}{R+r}$$$$\lambda =\frac{r}{R}$$$$\Rightarrow \frac{\lambda }{1-\lambda }=\frac{2\sqrt{\lambda }}{1+\lambda }$$$$\Rightarrow \lambda (1+\lambda )=2(1-\lambda )\sqrt{\lambda }$$$$\Rightarrow {\lambda }^2{(1+\lambda )}^2=4\lambda {(1-\lambda )}^2$$$$\Rightarrow \lambda {(1+\lambda )}^2=4{(1-\lambda )}^2$$$$\Rightarrow {\lambda }^3+2{\lambda }^2+\lambda =4{\lambda }^2-8\lambda +4$$$$\Rightarrow {\lambda }^3-2{\lambda }^2+9\lambda -4=0$$$$\Rightarrow \lambda =0.48389$$

Commented by ajfour last updated on 16/Dec/18

$$ThanksforsolvingSir.$$

Commented by tanmay.chaudhury50@gmail.com last updated on 17/Dec/18

$$sirplsexplainthdfirstlinepls...$$

Commented by mr W last updated on 17/Dec/18

Commented by mr W last updated on 17/Dec/18

$$\frac{BD}{BC}=\frac{AE}{AC}$$$$\Rightarrow \frac{r}{R-r}=\frac{\sqrt{{(R+r)}^2-{(R-r)}^2}}{R+r}$$

Commented by tanmay.chaudhury50@gmail.com last updated on 17/Dec/18

$$excellentthankyousir...$$

Commented by ajfour last updated on 17/Dec/18

$$\cos \mathrm{\angle }DBC=\cos \theta =\frac{DB}{BC}=\frac{r}{R-r}$$$$\sin \mathrm{\angle }CAE=\sin \theta =\frac{CE}{CA}=\frac{R-r}{R+r}$$$$hence{\left(\frac{r}{R-r}\right)}^2+{\left(\frac{R-r}{R+r}\right)}^2=1$$$$letx=\frac{r}{R}$$$$\Rightarrow {\left(\frac{x}{1-x}\right)}^2+{\left(\frac{1-x}{1+x}\right)}^2=1$$$$\Rightarrow x^2{(1+x)}^2+{(1-x)}^4={(1-x^2)}^2$$$$x^4+2x^3+x^2+1+x^4-4x+6x^2-4x^3$$$$=1+x^4-2x^2$$$$\Rightarrow x^4-2x^3+9x^2-4x=0$$$$ifx\ne 0,$$$$x^3-2x^2+9x-4=0$$

Commented by tanmay.chaudhury50@gmail.com last updated on 17/Dec/18

$$thankyousir...$$

Answered by ajfour last updated on 16/Dec/18

$$igetr=0.48389R$$

Commented by peter frank last updated on 16/Dec/18

$$pleasehelpQN50349$$

Commented by ajfour last updated on 16/Dec/18

$$thanksSir!forconfirming.$$

Commented by MJS last updated on 16/Dec/18

$$\mathrm{me}\mathrm{too}.$$$${\mathrm{I}}^{\prime }\mathrm{ve}\mathrm{been}\mathrm{trying}\mathrm{with}\mathrm{an}\mathrm{ellipse}\mathrm{instead}\mathrm{of}$$$$\mathrm{the}\mathrm{big}\mathrm{circle},\mathrm{it}\mathrm{seems}\mathrm{very}\mathrm{hard}...\mathrm{not}\mathrm{sure}$$$$\mathrm{if}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{possible}\mathrm{to}\mathrm{solve}.$$

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