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Question Number 103756 by ajfour last updated on 17/Jul/20

Commented by ajfour last updated on 17/Jul/20

Find radius r of shown circle.

$${Find}\:{radius}\:{r}\:{of}\:{shown}\:{circle}. \\ $$

Answered by mr W last updated on 17/Jul/20

center of circle (h,r) y=x^2 tan θ=y′=2x P(p,p^2 ) p=h−r sin θ p^2 =r+r cos θ tan θ=2p p=h−r ((2p)/(√(1+4p^2 ))) p^2 =r(1+ (1/(√(1+4p^2 )))) { ((h=p(1+((2p^2 )/(1+(√(1+4p^2 ))))))),((r=(p^2 /(1+(1/(√(1+4p^2 ))))))) :} ...(I) y=(√x) tan ϕ=y′=(1/(2(√x))) Q(q^2 ,q) q^2 =h−r sin ϕ q=r+r cos ϕ tan ϕ=(1/(2q)) q^2 =h−r (1/(√(1+4q^2 ))) q=r(1+((2q)/(√(1+4q^2 )))) { ((h=q(q+(1/(2q+(√(1+4q^2 ))))))),((r=(q/(1+((2q)/(√(1+4q^2 ))))))) :} ...(II) intersection point of curve (I) & (II): h≈1.4167 r≈0.56886

$${center}\:{of}\:{circle}\:\left({h},{r}\right) \\ $$$$ \\ $$$${y}={x}^{\mathrm{2}} \\ $$$$\mathrm{tan}\:\theta={y}'=\mathrm{2}{x} \\ $$$${P}\left({p},{p}^{\mathrm{2}} \right) \\ $$$${p}={h}−{r}\:\mathrm{sin}\:\theta \\ $$$${p}^{\mathrm{2}} ={r}+{r}\:\mathrm{cos}\:\theta \\ $$$$\mathrm{tan}\:\theta=\mathrm{2}{p} \\ $$$${p}={h}−{r}\:\frac{\mathrm{2}{p}}{\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }} \\ $$$${p}^{\mathrm{2}} ={r}\left(\mathrm{1}+\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }}\right) \\ $$$$\begin{cases}{{h}={p}\left(\mathrm{1}+\frac{\mathrm{2}{p}^{\mathrm{2}} }{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }}\right)}\\{{r}=\frac{{p}^{\mathrm{2}} }{\mathrm{1}+\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }}}}\end{cases}\:\:\:\:\:...\left({I}\right) \\ $$$$ \\ $$$${y}=\sqrt{{x}} \\ $$$$\mathrm{tan}\:\varphi={y}'=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}} \\ $$$${Q}\left({q}^{\mathrm{2}} ,{q}\right) \\ $$$${q}^{\mathrm{2}} ={h}−{r}\:\mathrm{sin}\:\varphi \\ $$$${q}={r}+{r}\:\mathrm{cos}\:\varphi \\ $$$$\mathrm{tan}\:\varphi=\frac{\mathrm{1}}{\mathrm{2}{q}} \\ $$$${q}^{\mathrm{2}} ={h}−{r}\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\mathrm{4}{q}^{\mathrm{2}} }} \\ $$$${q}={r}\left(\mathrm{1}+\frac{\mathrm{2}{q}}{\sqrt{\mathrm{1}+\mathrm{4}{q}^{\mathrm{2}} }}\right) \\ $$$$\begin{cases}{{h}={q}\left({q}+\frac{\mathrm{1}}{\mathrm{2}{q}+\sqrt{\mathrm{1}+\mathrm{4}{q}^{\mathrm{2}} }}\right)}\\{{r}=\frac{{q}}{\mathrm{1}+\frac{\mathrm{2}{q}}{\sqrt{\mathrm{1}+\mathrm{4}{q}^{\mathrm{2}} }}}}\end{cases}\:\:\:...\left({II}\right) \\ $$$${intersection}\:{point}\:{of}\:{curve}\:\left({I}\right)\:\&\:\left({II}\right): \\ $$$${h}\approx\mathrm{1}.\mathrm{4167} \\ $$$${r}\approx\mathrm{0}.\mathrm{56886} \\ $$

Commented by mr W last updated on 17/Jul/20

Commented by mr W last updated on 17/Jul/20

Commented by ajfour last updated on 17/Jul/20

parametric way! Anyway thanks Sir, i can barely think of any other way out...

$${parametric}\:{way}!\:\:{Anyway}\:{thanks} \\ $$$${Sir},\:\:{i}\:{can}\:{barely}\:{think}\:{of}\:{any} \\ $$$${other}\:{way}\:{out}... \\ $$

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