Question-203232

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

center of circle at C(h, k) radius of circle r y=x−x^3 y′=1−3x^2 at point O(0, 0): tan ϕ=y′=1 ⇒ϕ=45° at point P(p, q): q=p−p^3 tan θ=−y′=3p^2 −1 h=r sin ϕ=(r/( (√2))) k=−r cos ϕ=−(r/( (√2))) h=p−r sin θ k=q−r cos θ ⇒p−r sin θ=(r/( (√2))) ⇒p=((1/( (√2)))+sin θ)r ⇒p−p^3 −r cos θ=−(r/( (√2))) ⇒p(1−p^2 )=(−(1/( (√2)))+cos θ)r 1−p^2 =((−1+(√2) cos θ)/(1+(√2) sin θ)) 1−p^2 =((−1+(√2)×(1/( (√(1+(3p^2 −1)^2 )))))/(1+(√2)×((3p^2 −1)/( (√(1+(3p^2 −1)^2 )))))) ⇒1−p^2 =(((√2)−(√(1+(3p^2 −1)^2 )))/( (√(1+(3p^2 −1)^2 ))+(√2)(3p^2 −1))) ⇒p≈1.1024 r=(p/((1/( (√2)))+((3p^2 −1)/( (√(1+(3p^2 −1)^2 ))))))≈0.6712 ✓

$${center}\:{of}\:{circle}\:{at}\:{C}\left({h},\:{k}\right) \\ $$$${radius}\:{of}\:{circle}\:{r} \\ $$$${y}={x}−{x}^{\mathrm{3}} \\ $$$${y}'=\mathrm{1}−\mathrm{3}{x}^{\mathrm{2}} \\ $$$${at}\:{point}\:{O}\left(\mathrm{0},\:\mathrm{0}\right): \\ $$$$\mathrm{tan}\:\varphi={y}'=\mathrm{1}\:\Rightarrow\varphi=\mathrm{45}° \\ $$$${at}\:{point}\:{P}\left({p},\:{q}\right): \\ $$$${q}={p}−{p}^{\mathrm{3}} \\ $$$$\mathrm{tan}\:\theta=−{y}'=\mathrm{3}{p}^{\mathrm{2}} −\mathrm{1} \\ $$$$ \\ $$$${h}={r}\:\mathrm{sin}\:\varphi=\frac{{r}}{\:\sqrt{\mathrm{2}}} \\ $$$${k}=−{r}\:\mathrm{cos}\:\varphi=−\frac{{r}}{\:\sqrt{\mathrm{2}}} \\ $$$${h}={p}−{r}\:\mathrm{sin}\:\theta \\ $$$${k}={q}−{r}\:\mathrm{cos}\:\theta \\ $$$$\Rightarrow{p}−{r}\:\mathrm{sin}\:\theta=\frac{{r}}{\:\sqrt{\mathrm{2}}}\:\Rightarrow{p}=\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\mathrm{sin}\:\theta\right){r} \\ $$$$\Rightarrow{p}−{p}^{\mathrm{3}} −{r}\:\mathrm{cos}\:\theta=−\frac{{r}}{\:\sqrt{\mathrm{2}}}\:\Rightarrow{p}\left(\mathrm{1}−{p}^{\mathrm{2}} \right)=\left(−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\mathrm{cos}\:\theta\right){r} \\ $$$$\mathrm{1}−{p}^{\mathrm{2}} =\frac{−\mathrm{1}+\sqrt{\mathrm{2}}\:\mathrm{cos}\:\theta}{\mathrm{1}+\sqrt{\mathrm{2}}\:\mathrm{sin}\:\theta} \\ $$$$\mathrm{1}−{p}^{\mathrm{2}} =\frac{−\mathrm{1}+\sqrt{\mathrm{2}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\left(\mathrm{3}{p}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }}}{\mathrm{1}+\sqrt{\mathrm{2}}×\frac{\mathrm{3}{p}^{\mathrm{2}} −\mathrm{1}}{\:\sqrt{\mathrm{1}+\left(\mathrm{3}{p}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }}} \\ $$$$\Rightarrow\mathrm{1}−{p}^{\mathrm{2}} =\frac{\sqrt{\mathrm{2}}−\sqrt{\mathrm{1}+\left(\mathrm{3}{p}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }}{\:\sqrt{\mathrm{1}+\left(\mathrm{3}{p}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }+\sqrt{\mathrm{2}}\left(\mathrm{3}{p}^{\mathrm{2}} −\mathrm{1}\right)} \\ $$$$\Rightarrow{p}\approx\mathrm{1}.\mathrm{1024} \\ $$$${r}=\frac{{p}}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{3}{p}^{\mathrm{2}} −\mathrm{1}}{\:\sqrt{\mathrm{1}+\left(\mathrm{3}{p}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }}}\approx\mathrm{0}.\mathrm{6712}\:\checkmark \\ $$

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