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Question Number 107696 by ajfour last updated on 12/Aug/20

Commented by ajfour last updated on 12/Aug/20

$$Ifbothcircleshavetheequalradius,$$$$findtheradius.$$

Answered by mr W last updated on 20/Aug/20

Commented by mr W last updated on 21/Aug/20

$$A(a,a^2)$$$$\tan \theta =y^{\prime }=2a$$$$B(b,b^2)$$$$\tan \phi =y^{\prime }=2b$$$$$$$$x_C=a-r\sin \theta =0$$$$y_C=a^2+r\cos \theta$$$$\Rightarrow a=\frac{\tan \theta }{2}=r\sin \theta$$$$\Rightarrow \cos \theta =\frac{1}{2r}$$$$y_C=\frac{{\tan }^2\theta }{4}+\frac{1}{2}=\frac{1-{\cos }^2\theta }{4{\cos }^2\theta }+\frac{1}{2}=r^2+\frac{1}{4}$$$$$$$$x_D=b-r\sin \phi =r$$$$y_D=b^2+r\cos \phi$$$$b=\frac{\tan \phi }{2}=r(1+\sin \phi )$$$$\Rightarrow r=\frac{\tan \phi }{2(1+\sin \phi )}$$$$y_D=r^2(1+{\sin }^2\phi +2\sin \phi )+r\cos \phi$$$${\left(2r\right)}^2-r^2={(y_D-y_C)}^2$$$$3r^2={[r^2\sin \phi (2+\sin \phi )+r\cos \phi -\frac{1}{4}]}^2$$$$\Rightarrow \frac{12{\tan }^2\phi }{{(1+\sin \phi )}^2}={[\frac{{\tan }^2\phi \sin \phi (2+\sin \phi )}{{(1+\sin \phi )}^2}+\frac{2\sin \phi }{1+\sin \phi }-1]}^2$$$$\Rightarrow \phi =1.1875=68.039°$$$$\Rightarrow r=0.6433$$

Commented by mr W last updated on 20/Aug/20

Commented by ajfour last updated on 21/Aug/20

$$ThankuSir,marvelloussolution!$$

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