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Question Number 42310 by ajfour last updated on 23/Aug/18

Commented by ajfour last updated on 23/Aug/18

$$SolutiontoQ.41766\left(Anothermethod\right)$$

Answered by ajfour last updated on 23/Aug/18

$$\cos \alpha =\frac{a}{2R};\cos \beta =\frac{b}{2R}$$$$\theta =\frac{\alpha +\beta }{2}$$$$eq.ofbiggercircle:$$$$x^2+y^2=R^2...\left(i\right)$$$$eq.ofsmallcircle:$$$${(x-h)}^2+{(y-k)}^2=r^2...\left(ii\right)$$$$with{\mathit{h}}^2+{\mathit{k}}^2={(\mathit{R}-\mathit{r})}^2...\left(iii\right)$$$$but\mathit{h}=\frac{\mathit{r}}{\sin \theta }\cos (\beta -\theta )-R$$$$and\mathit{k}=-\frac{\mathit{r}}{\sin \theta }\sin (\beta -\theta )$$$$substitutingthesein\left(iii\right)wehave$$$$\frac{r^2}{\sin {}^2\theta }-\frac{2rR}{\sin \theta }\cos (\beta -\theta )=r^2-2rR$$$$\Rightarrow r=0or$$$$\mathit{r}=\frac{2R[\frac{\cos (\beta -\theta )}{\sin \theta }-1]}{\frac{1}{\sin {}^2\theta }-1}$$$$but\theta =\frac{\alpha +\beta }{2},hence$$$$\mathit{r}=2\mathit{R}[\frac{\cos \left(\frac{\beta -\alpha }{2}\right)}{\cos \left(\frac{\beta +\alpha }{2}\right)}-\tan \left(\frac{\beta +\alpha }{2}\right)]\tan \left(\frac{\beta +\alpha }{2}\right);$$$$with\beta ={\cos }^{-1}\frac{b}{2R}\mathrm{\&}\alpha ={\cos }^{-1}\frac{a}{2R}.$$

Commented by MrW3 last updated on 24/Aug/18

$$verynice!theansweriscorrect.$$

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