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Question Number 104157 by ajfour last updated on 19/Jul/20

Commented by ajfour last updated on 19/Jul/20

$$FindradiiRandrofbiggerand$$$$smallcircles.$$

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

$$H(1,1)$$$$OH=\sqrt{2}R+R=\sqrt{2}$$$$\Rightarrow R=\frac{\sqrt{2}}{\sqrt{2}+1}=2-\sqrt{2}$$$$D(R+2\sqrt{Rr},r)=(h,r)$$$$F(k,\frac{1}{k})$$$$\tan \theta =\frac{1}{k^2}$$$$k=h+r\sin \theta =R+2\sqrt{Rr}+r\frac{1}{\sqrt{1+k^4}}$$$$\frac{1}{k}=r+r\cos \theta =r(1+\frac{k^2}{\sqrt{1+k^4}})$$$$\Rightarrow r=\frac{1}{k(1+\frac{k^2}{\sqrt{1+k^4}})}$$$$k=R+2\sqrt{\frac{R}{k(1+\frac{k^2}{\sqrt{1+k^4}})}}+\frac{1}{k(\sqrt{1+k^4}+k^2)}$$$$\Rightarrow k\approx 1.5831$$$$\Rightarrow r\approx 0.3275$$

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

Commented by ajfour last updated on 19/Jul/20

$$Itoothink,thisasthepossibleway,$$$$Sir,thanksforthesolution!$$

Answered by ajfour last updated on 21/Jul/20

Commented by ajfour last updated on 22/Jul/20

$$R=2-\sqrt{2}$$$$C(0,R\sqrt{2})$$$$lety-x=s,y+x=t$$$$\Rightarrow y=\frac{s+t}{2},\frac{1}{x}=\frac{2}{t-s}$$$$y=\frac{1}{x}is{t}^2-s^2=4$$$$s=\sqrt{t^2-4}$$$$A(\frac{R}{\sqrt{2}}+\sqrt{2Rr},\frac{R}{\sqrt{2}}+\sqrt{2Rr})$$$$D(\frac{R}{\sqrt{2}}+\sqrt{2Rr}-\frac{r}{\sqrt{2}},\frac{R}{\sqrt{2}}+\sqrt{2Rr}+\frac{r}{\sqrt{2}})$$$$.....$$$$$$

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