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Question Number 182298 by SEKRET last updated on 07/Dec/22

Answered by mr W last updated on 07/Dec/22

radius: r  center: (k,2−r)  touching point with y=x^2 : (p,p^2 )  touching point with y=(x^2 /4): (q,(q^2 /4))  tan θ=2p  k=p+r sin θ  ⇒k=p+((2pr)/( (√(1+4p^2 ))))  2−r=p^2 −r cos θ  ⇒2−r=p^2 −(r/( (√(1+4p^2 ))))  tan ϕ=(q/2)  k=q−r sin ϕ  ⇒k=q−((qr)/( (√(4+q^2 ))))  2−r=(q^2 /4)+r cos ϕ  ⇒2−r=(q^2 /4)+((2r)/( (√(4+q^2 ))))  ⇒r=((q−p)/(((2p)/( (√(1+4p^2 ))))+(q/( (√(4+q^2 ))))))  ⇒r=((2−p^2 )/(1−(1/( (√(1+4p^2 ))))))  ⇒r=((2−(q^2 /4))/(1+(2/( (√(4+q^2 ))))))  ⇒p≈1.29415, q≈2.13965  ⇒r≈0.5084

$${radius}:\:{r} \\ $$$${center}:\:\left({k},\mathrm{2}−{r}\right) \\ $$$${touching}\:{point}\:{with}\:{y}={x}^{\mathrm{2}} :\:\left({p},{p}^{\mathrm{2}} \right) \\ $$$${touching}\:{point}\:{with}\:{y}=\frac{{x}^{\mathrm{2}} }{\mathrm{4}}:\:\left({q},\frac{{q}^{\mathrm{2}} }{\mathrm{4}}\right) \\ $$$$\mathrm{tan}\:\theta=\mathrm{2}{p} \\ $$$${k}={p}+{r}\:\mathrm{sin}\:\theta \\ $$$$\Rightarrow{k}={p}+\frac{\mathrm{2}{pr}}{\:\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }} \\ $$$$\mathrm{2}−{r}={p}^{\mathrm{2}} −{r}\:\mathrm{cos}\:\theta \\ $$$$\Rightarrow\mathrm{2}−{r}={p}^{\mathrm{2}} −\frac{{r}}{\:\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }} \\ $$$$\mathrm{tan}\:\varphi=\frac{{q}}{\mathrm{2}} \\ $$$${k}={q}−{r}\:\mathrm{sin}\:\varphi \\ $$$$\Rightarrow{k}={q}−\frac{{qr}}{\:\sqrt{\mathrm{4}+{q}^{\mathrm{2}} }} \\ $$$$\mathrm{2}−{r}=\frac{{q}^{\mathrm{2}} }{\mathrm{4}}+{r}\:\mathrm{cos}\:\varphi \\ $$$$\Rightarrow\mathrm{2}−{r}=\frac{{q}^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{2}{r}}{\:\sqrt{\mathrm{4}+{q}^{\mathrm{2}} }} \\ $$$$\Rightarrow{r}=\frac{{q}−{p}}{\frac{\mathrm{2}{p}}{\:\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }}+\frac{{q}}{\:\sqrt{\mathrm{4}+{q}^{\mathrm{2}} }}} \\ $$$$\Rightarrow{r}=\frac{\mathrm{2}−{p}^{\mathrm{2}} }{\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }}} \\ $$$$\Rightarrow{r}=\frac{\mathrm{2}−\frac{{q}^{\mathrm{2}} }{\mathrm{4}}}{\mathrm{1}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{4}+{q}^{\mathrm{2}} }}} \\ $$$$\Rightarrow{p}\approx\mathrm{1}.\mathrm{29415},\:{q}\approx\mathrm{2}.\mathrm{13965} \\ $$$$\Rightarrow{r}\approx\mathrm{0}.\mathrm{5084} \\ $$

Commented by mr W last updated on 07/Dec/22

Commented by SEKRET last updated on 07/Dec/22

 thank  you

$$\:\boldsymbol{\mathrm{thank}}\:\:\boldsymbol{\mathrm{you}}\: \\ $$

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