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Question Number 126192 by ajfour last updated on 18/Dec/20

Commented by ajfour last updated on 18/Dec/20

$$Ifshapeofparabolaissameas$$$$thatofy=x^2,findradiusof$$$$thesmallercircle.$$

Commented by MJS_new last updated on 19/Dec/20

$$\mathrm{I}\mathrm{get}$$$$r_A\approx .196437$$$$r_B\approx .344870$$$$\mathrm{par}:y=x^2+.365017$$

Commented by MJS_new last updated on 19/Dec/20

$$\mathrm{I}\mathrm{corrected}\mathrm{my}\mathrm{values}\mathrm{but}{\mathrm{they}}^{\prime }\mathrm{re}\mathrm{still}\mathrm{not}$$$$\mathrm{the}\mathrm{same}\mathrm{as}\mathrm{yours}$$

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

$$nowtheyarethesame,seeupdate$$$$inmysolution.$$

Commented by MJS_new last updated on 20/Dec/20

$$\mathrm{yes}.$$

Answered by mr W last updated on 18/Dec/20

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

$$y=x^2+k$$$$P(p,k+p^2)$$$$\tan \theta =2p$$$$A(r,r)$$$$p=r-r\sin \theta$$$$\Rightarrow r\sin \theta =r-p$$$$p^2+k=r+r\cos \theta$$$$\Rightarrow r\cos \theta =p^2+k-r$$$$\tan \theta =\frac{r-p}{p^2+k-r}=2p$$$${(r-p)}^2+{(p^2+k-r)}^2=r^2$$$${(r-p)}^2(1+4p^2)=4r^2{p}^2$$$$r^2-2p(1+4p^2)r+p^2(1+4p^2)=0$$$$\Rightarrow r=r\left(p\right)=p(1+4p^2)+2p^2\sqrt{1+4p^2}...\left(i\right)$$$$\Rightarrow k=k\left(p\right)=\frac{r-p}{2p}+r-p^2...\left(ii\right)$$$$$$$$Q(q,k+q^2)$$$$\tan \phi =2q$$$$B(r+2\sqrt{Rr},R)$$$$q=r+2\sqrt{Rr}-R\sin \phi$$$$\Rightarrow R\sin \phi =r+2\sqrt{Rr}-q$$$$q^2+k=R+R\cos \phi$$$$\Rightarrow R\cos \phi =q^2+k-R$$$$\tan \phi =\frac{r+2\sqrt{Rr}-q}{q^2+k-R}=2q$$$$r+2\sqrt{Rr}-q=2q(q^2+k-R)...\left(I\right)$$$$\Rightarrow q^3-(R-k-\frac{1}{2})q-(\frac{r}{2}+\sqrt{Rr})=0$$$$\Rightarrow q=\sqrt[3]{\sqrt{\frac{1}{4}{(\frac{r}{2}+\sqrt{Rr})}^2-\frac{1}{27}{(R-k-\frac{1}{2})}^3}+\frac{1}{2}(\frac{r}{2}+\sqrt{Rr})}$$$$-\sqrt[3]{\sqrt{\frac{1}{4}{(\frac{r}{2}+\sqrt{Rr})}^2-\frac{1}{27}{(R-k-\frac{1}{2})}^3}-\frac{1}{2}(\frac{r}{2}+\sqrt{Rr})}$$$$$$$${(r+2\sqrt{Rr}-q)}^2+{(q^2+k-R)}^2=R^2$$$$(1+4q^2){(q^2+k-R)}^2=R^2...\left(II\right)$$$$$$$$\mathit{g}\mathit{e}\mathit{n}\mathit{e}\mathit{r}\mathit{a}\mathit{l}\mathit{c}\mathit{a}\mathit{s}\mathit{e}:$$$$foranyRwecangetpfrom\left(I\right)and$$$$\left(II\right),thenr\mathrm{\&}kfrom\left(i\right)and\left(ii\right).$$$$Example:$$$$R=10$$$$\Rightarrow p\approx 0.59646$$$$\Rightarrow r\approx 2.5528,k\approx 3.8371$$$$$$$$\mathit{s}\mathit{p}\mathit{e}\mathit{c}\mathit{i}\mathit{a}\mathit{l}\mathit{c}\mathit{a}\mathit{s}\mathit{e}\mathit{A}\mathit{B}\mathrm{\perp }\mathit{A}\mathit{C}:$$$$\tan \theta =\frac{R-r}{2\sqrt{Rr}}=2p$$$$R-4p\sqrt{r}\sqrt{R}-r=0$$$$\sqrt{R}=(2p+\sqrt{1+4p^2})\sqrt{r}$$$$\Rightarrow R={(2p+\sqrt{1+4p^2})}^{2}r$$$$putthisinto\left(I\right)and\left(II\right)andweget$$$$p\approx 0.1426$$$$\Rightarrow r\approx 0.1965$$$$\Rightarrow k\approx 0.3651$$$$\Rightarrow R\approx 0.3450$$

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

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

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

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