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Question Number 203232 by Mathstar last updated on 13/Jan/24

Answered by mr W last updated on 13/Jan/24

Commented by mr W last updated on 13/Jan/24

$$centerofcircleatC(h,k)$$$$radiusofcircler$$$$y=x-x^3$$$$y^{\prime }=1-3x^2$$$$atpointO(0,0):$$$$\tan \phi =y^{\prime }=1\Rightarrow \phi =45°$$$$atpointP(p,q):$$$$q=p-p^3$$$$\tan \theta =-y^{\prime }=3p^2-1$$$$$$$$h=r\sin \phi =\frac{r}{\sqrt{2}}$$$$k=-r\cos \phi =-\frac{r}{\sqrt{2}}$$$$h=p-r\sin \theta$$$$k=q-r\cos \theta$$$$\Rightarrow p-r\sin \theta =\frac{r}{\sqrt{2}}\Rightarrow p=(\frac{1}{\sqrt{2}}+\sin \theta )r$$$$\Rightarrow p-p^3-r\cos \theta =-\frac{r}{\sqrt{2}}\Rightarrow p(1-p^2)=(-\frac{1}{\sqrt{2}}+\cos \theta )r$$$$1-p^2=\frac{-1+\sqrt{2}\cos \theta }{1+\sqrt{2}\sin \theta }$$$$1-p^2=\frac{-1+\sqrt{2}\times \frac{1}{\sqrt{1+{(3p^2-1)}^2}}}{1+\sqrt{2}\times \frac{3p^2-1}{\sqrt{1+{(3p^2-1)}^2}}}$$$$\Rightarrow 1-p^2=\frac{\sqrt{2}-\sqrt{1+{(3p^2-1)}^2}}{\sqrt{1+{(3p^2-1)}^2}+\sqrt{2}(3p^2-1)}$$$$\Rightarrow p\approx 1.1024$$$$r=\frac{p}{\frac{1}{\sqrt{2}}+\frac{3p^2-1}{\sqrt{1+{(3p^2-1)}^2}}}\approx 0.6712✓$$

Commented by mr W last updated on 13/Jan/24

Commented by mr W last updated on 13/Jan/24

Commented by Mathstar last updated on 13/Jan/24

Thank you ��

Commented by MathematicalUser2357 last updated on 20/Jan/24

$$r=\frac{4\sqrt{3+3\sqrt{7}}}{6\sqrt{2}+3\sqrt{14}}$$

Commented by mr W last updated on 20/Jan/24

$$i{havn}^{\prime }ttriedtogettheexactsolution.$$$$canyoushowhowyousolved?$$

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