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Question Number 41766 by ajfour last updated on 12/Aug/18

Answered by MrW3 last updated on 18/Aug/18

$${let}^{\prime }ssay0<b⩽a⩽2R$$$$$$$$B(0,0)$$$$O(\frac{b}{2},\sqrt{R^2-\frac{b^2}{4}})$$$$C(h,r)$$$$$$$$Eqn.ofBA:$$$$y=mxwithm=\tan ({\cos }^{-1}\frac{b}{2R}\pm {\cos }^{-1}\frac{a}{2R})$$$$Eqn.ofcirclewithr:$$$${(x-h)}^2+{(y-r)}^2=r^2$$$$\Rightarrow (x-h)+(y-r)\frac{dy}{dx}=0$$$$atM:$$$$(x_M-h)+(y_M-r)m=0$$$${(x_M-h)}^2+{(y_M-r)}^2=r^2$$$$\Rightarrow (m^2+1){(x_M-h)}^2=m^2{r}^2$$$$\Rightarrow x_M=h\pm \frac{mr}{\sqrt{m^2+1}}$$$$\Rightarrow y_M=r\mp \frac{r}{\sqrt{m^2+1}}$$$$y_M={mx}_M$$$$\Rightarrow y_M=r\mp \frac{r}{\sqrt{m^2+1}}=mh\pm \frac{m^{2}r}{\sqrt{m^2+1}}$$$$\Rightarrow 1\mp \frac{1}{\sqrt{m^2+1}}=\frac{mh}{r}\pm \frac{m^2}{\sqrt{m^2+1}}$$$$\Rightarrow h=\frac{r}{m}(1\pm \sqrt{m^2+1})=\lambda r>0$$$$with\lambda =\frac{1}{m}(1+\sqrt{m^2+1})$$$$$$$$OC=R-r$$$$\sqrt{{(h-\frac{b}{2})}^2+{(r-\sqrt{R^2-\frac{b^2}{4}})}^2}=R-r$$$${(\lambda r-\frac{b}{2})}^2+{(r-\sqrt{R^2-\frac{b^2}{4}})}^2={(R-r)}^2$$$${\lambda }^2{r}^2+\frac{b^2}{4}-\lambda br+r^2+R^2-\frac{b^2}{4}-2r\sqrt{R^2-\frac{b^2}{4}}=R^2+r^2-2Rr$$$${\lambda }^{2}r=\lambda b+2\sqrt{R^2-\frac{b^2}{4}}-2R$$$$\Rightarrow r=\frac{1}{{\lambda }^2}[\lambda b-2(R-\sqrt{R^2-\frac{b^2}{4}})]$$$$with\lambda =\frac{1}{m}(1+\sqrt{m^2+1})$$$$andm=\tan ({\cos }^{-1}\frac{b}{2R}\pm {\cos }^{-1}\frac{a}{2R})$$$$=\frac{\sqrt{1-\frac{b^2}{4R^2}}\times \frac{a}{2R}\pm \frac{b}{2R}\times \sqrt{1-\frac{a^2}{4R^2}}}{\frac{b}{2R}\times \frac{a}{2R}\mp \sqrt{1-\frac{b^2}{4R^2}}\times \sqrt{1-\frac{a^2}{4R^2}}}$$$$=\frac{a\sqrt{4R^2-b^2}\pm b\sqrt{4R^2-a^2}}{ab\mp \sqrt{(4R^2-a^2)(4R^2-b^2)}}$$$$$$$$with\alpha =\frac{a}{2R}⩽1,\beta =\frac{b}{2R}⩽1and\beta ⩽\alpha ,$$$$m=\frac{\alpha \sqrt{1-{\beta }^2}\pm \beta \sqrt{1-{\alpha }^2}}{\alpha \beta \mp \sqrt{(1-{\alpha }^2)(1-{\beta }^2)}}$$$$\lambda =\frac{1}{m}(1+\sqrt{m^2+1})$$$$\Rightarrow r=\frac{2R}{{\lambda }^2}(\lambda \beta +\sqrt{1-{\beta }^2}-1)$$

Commented by ajfour last updated on 23/Aug/18

$$Thankyousomuchsir!$$

Commented by MrW3 last updated on 13/Aug/18

Commented by MrW3 last updated on 14/Aug/18

$$anyothermethod?$$

Commented by ajfour last updated on 24/Aug/18

$$SeeQ.42310Sir..$$

Answered by ajfour last updated on 23/Aug/18

$$r=2R[\frac{\cos \left(\frac{\beta -\alpha }{2}\right)}{\cos \left(\frac{\alpha +\beta }{2}\right)}-\tan \left(\frac{\alpha +\beta }{2}\right)]\tan \left(\frac{\alpha +\beta }{2}\right)$$$$\mathit{w}\mathit{i}\mathit{t}\mathit{h}\cos \alpha =\frac{a}{2R},\cos \beta =\frac{b}{2R}.$$$$$$$$pleasehelpincheckingthisanswer$$$$Sir!$$

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