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Question Number 105422 by ajfour last updated on 28/Jul/20

Commented by ajfour last updated on 28/Jul/20

$$Findradiusofcircleintermsof$$$$aandb.$$

Answered by ajfour last updated on 28/Jul/20

$$letA(a-a\cos \theta ,b+b\sin \theta )$$$$slopeofAC=m=-\frac{a\sin \theta }{b\cos \theta }$$$$\underset{-}{Equations\left(i\right)\mathrm{\&}\left(ii\right):}$$$$x_C=a-a\cos \theta -r\left(\frac{b\cos \theta }{\sqrt{a^2\sin {}^2\theta +b^2\cos {}^2\theta }}\right)$$$$y_C=b+b\sin \theta +r\left(\frac{a\sin \theta }{\sqrt{a^2\sin {}^2\theta +b^2\cos {}^2\theta }}\right)$$$$Also{x}_C=r....\left(iii\right)$$$$Andlet{x}_B=r+r\cos \varphi ...\left(iv\right)$$$$y_B=y_C+r\sin \varphi ....\left(v\right)$$$$y_B=x_B^2...\left(vi\right)\tan \varphi =\frac{1}{2x_B}..\left(vii\right)$$$$\Rightarrow \frac{1}{4\tan {}^2\varphi }=y_C+r\sin \varphi$$$$\Rightarrow \frac{1}{4\tan {}^2\varphi }-r\sin \varphi =$$$$b+b\sin \theta +r\left(\frac{a\sin \theta }{\sqrt{a^2\sin {}^2\theta +b^2\cos {}^2\theta }}\right)$$$$Andfrom\left(iii\right):$$$$r=a-a\cos \theta -r\left(\frac{b\cos \theta }{\sqrt{a^2\sin {}^2\theta +b^2\cos {}^2\theta }}\right)$$$$And\frac{1}{2\tan \varphi }=r+r\cos \varphi$$$$\Rightarrow r=\frac{1}{2\tan \varphi (1+\cos \varphi )}$$$$.....$$

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