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Question Number 225716 by ajfour last updated on 07/Nov/25
Commented by ajfour last updated on 07/Nov/25
Find radius of blue circle R.
$${Find}\:{radius}\:{of}\:{blue}\:{circle}\:{R}. \\ $$
Commented by fantastic last updated on 07/Nov/25
https://youtu.be/0wyMmkWtgTQ?si=tdjjg8E448DLBJEU
Commented by fantastic last updated on 07/Nov/25
Commented by fantastic last updated on 07/Nov/25
α=sin^(−1) (((R−r)/(R+r))) x=(√((R+r)^2 −(R−r)^2 ))+(r/(tan α)) =2(√(Rr))+(r/((sin α)/(cos α))) [cos α=(√(1−sin^2 α))=(√((4Rr)/((R+r)^2 )))=(2/((R+r)))(√(Rr))] ∴tan α=(((R−r)/(R+r))/((2/((R+r)))(√(Rr))))=((R−r)/(2(√(Rr)))) x=2(√(Rr))(1+(r/(R−r)))=2R(√(Rr))/R−r Base=((4R(√(Rr)))/(R−r)) h=(x/(tan 2α)) a=(√((h−R)^2 −R^2 )) Two sides=a+x
$$\alpha=\mathrm{sin}^{−\mathrm{1}} \left(\frac{{R}−{r}}{{R}+{r}}\right) \\ $$$${x}=\sqrt{\left({R}+{r}\right)^{\mathrm{2}} −\left({R}−{r}\right)^{\mathrm{2}} }+\frac{{r}}{\mathrm{tan}\:\alpha} \\ $$$$=\mathrm{2}\sqrt{{Rr}}+\frac{{r}}{\frac{\mathrm{sin}\:\alpha}{\mathrm{cos}\:\alpha}} \\ $$$$\left[\mathrm{cos}\:\alpha=\sqrt{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \alpha}=\sqrt{\frac{\mathrm{4}{Rr}}{\left({R}+{r}\right)^{\mathrm{2}} }}=\frac{\mathrm{2}}{\left({R}+{r}\right)}\sqrt{{Rr}}\right] \\ $$$$\therefore\mathrm{tan}\:\alpha=\frac{\frac{{R}−{r}}{{R}+{r}}}{\frac{\mathrm{2}}{\left({R}+{r}\right)}\sqrt{{Rr}}}=\frac{{R}−{r}}{\mathrm{2}\sqrt{{Rr}}} \\ $$$${x}=\mathrm{2}\sqrt{{Rr}}\left(\mathrm{1}+\frac{{r}}{{R}−{r}}\right)=\mathrm{2}{R}\sqrt{{Rr}}/{R}−{r} \\ $$$${Base}=\frac{\mathrm{4}{R}\sqrt{{Rr}}}{{R}−{r}} \\ $$$${h}=\frac{{x}}{\mathrm{tan}\:\mathrm{2}\alpha} \\ $$$${a}=\sqrt{\left({h}−{R}\right)^{\mathrm{2}} −{R}^{\mathrm{2}} } \\ $$$${Two}\:{sides}={a}+{x} \\ $$
Answered by mr W last updated on 09/Nov/25
both parabolas have the same shape. due to symmetry, the center of red circle lies at ((1/4), ((17)/(36))). say its radius is r. then its equation is (x−(1/4))^2 +(y−((17)/(36)))^2 =r^2 . say it touches the parabola y=(1/4)−x^2 at (ξ, (1/4)−ξ^2 ) (ξ−(1/4))^2 +((1/4)−ξ^2 −((17)/(36)))^2 =r^2 (1/(2ξ))=(((1/4)−ξ^2 −((17)/(36)))/(ξ−(1/4))) ξ^3 +((13ξ)/(18))−(1/8)=0 ξ=((((29(√(321)))/(3888))+(1/(16))))^(1/3) −((((29(√(321)))/(3888))−(1/(16))))^(1/3)
$${both}\:{parabolas}\:{have}\:{the}\:{same}\:{shape}. \\ $$$${due}\:{to}\:{symmetry},\:{the}\:{center}\:{of}\:{red} \\ $$$${circle}\:{lies}\:{at}\:\left(\frac{\mathrm{1}}{\mathrm{4}},\:\frac{\mathrm{17}}{\mathrm{36}}\right).\:{say}\:{its}\:{radius} \\ $$$${is}\:{r}.\:{then}\:{its}\:{equation}\:{is} \\ $$$$\left({x}−\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} +\left({y}−\frac{\mathrm{17}}{\mathrm{36}}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} . \\ $$$${say}\:{it}\:{touches}\:{the}\:{parabola}\:{y}=\frac{\mathrm{1}}{\mathrm{4}}−{x}^{\mathrm{2}} \\ $$$${at}\:\left(\xi,\:\frac{\mathrm{1}}{\mathrm{4}}−\xi^{\mathrm{2}} \right) \\ $$$$\left(\xi−\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{4}}−\xi^{\mathrm{2}} −\frac{\mathrm{17}}{\mathrm{36}}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\xi}=\frac{\frac{\mathrm{1}}{\mathrm{4}}−\xi^{\mathrm{2}} −\frac{\mathrm{17}}{\mathrm{36}}}{\xi−\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$\xi^{\mathrm{3}} +\frac{\mathrm{13}\xi}{\mathrm{18}}−\frac{\mathrm{1}}{\mathrm{8}}=\mathrm{0} \\ $$$$\xi=\sqrt[{\mathrm{3}}]{\frac{\mathrm{29}\sqrt{\mathrm{321}}}{\mathrm{3888}}+\frac{\mathrm{1}}{\mathrm{16}}}−\sqrt[{\mathrm{3}}]{\frac{\mathrm{29}\sqrt{\mathrm{321}}}{\mathrm{3888}}−\frac{\mathrm{1}}{\mathrm{16}}} \\ $$
Commented by mr W last updated on 09/Nov/25
Commented by mr W last updated on 09/Nov/25
Commented by mr W last updated on 09/Nov/25
say the blue circle with radius R touches the parabola y=(1/4)−x^2 at (p, (1/4)−p^2 ) tan φ=(1/(2p)) h=p+R cos φ=p+((2pR)/( (√(1+4p^2 )))) k=(1/4)−p^2 +R sin φ=(1/4)−p^2 +(R/( (√(1+4p^2 )))) (p+((2pR)/( (√(1+4p^2 ))))−(1/4))^2 +((1/4)−p^2 +(R/( (√(1+4p^2 ))))−((17)/(36)))^2 =(r+R)^2 say blue circle touches the parabola y=((25)/(36))+(x−(1/2))^2 at (q, ((25)/(36))+(q−(1/2))). tan ϕ=2(q−(1/2))=2q−1 h=q+R sin ϕ=q+(((2q−1)R)/( (√(1+(2q−1)^2 )))) k=((25)/(36))+(q−(1/2))^2 +R cos ϕ=((25)/(36))+(q−(1/2))^2 +(R/( (√(1+(2q−1)^2 )))) q+(((2q−1)R)/( (√(1+(2q−1)^2 ))))=p+((2pR)/( (√(1+4p^2 )))) ((25)/(36))+(q−(1/2))^2 +(R/( (√(1+(2q−1)^2 ))))=(1/4)−p^2 +(R/( (√(1+4p^2 ))))
$${say}\:{the}\:{blue}\:{circle}\:{with}\:{radius}\:{R} \\ $$$${touches}\:{the}\:{parabola}\:{y}=\frac{\mathrm{1}}{\mathrm{4}}−{x}^{\mathrm{2}} \:{at} \\ $$$$\left({p},\:\frac{\mathrm{1}}{\mathrm{4}}−{p}^{\mathrm{2}} \right) \\ $$$$\mathrm{tan}\:\phi=\frac{\mathrm{1}}{\mathrm{2}{p}} \\ $$$${h}={p}+{R}\:\mathrm{cos}\:\phi={p}+\frac{\mathrm{2}{pR}}{\:\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }} \\ $$$${k}=\frac{\mathrm{1}}{\mathrm{4}}−{p}^{\mathrm{2}} +{R}\:\mathrm{sin}\:\phi=\frac{\mathrm{1}}{\mathrm{4}}−{p}^{\mathrm{2}} +\frac{{R}}{\:\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }} \\ $$$$\left({p}+\frac{\mathrm{2}{pR}}{\:\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }}−\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{4}}−{p}^{\mathrm{2}} +\frac{{R}}{\:\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }}−\frac{\mathrm{17}}{\mathrm{36}}\right)^{\mathrm{2}} =\left({r}+{R}\right)^{\mathrm{2}} \\ $$$${say}\:{blue}\:{circle}\:{touches}\:{the}\:{parabola} \\ $$$${y}=\frac{\mathrm{25}}{\mathrm{36}}+\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:{at}\:\left({q},\:\frac{\mathrm{25}}{\mathrm{36}}+\left({q}−\frac{\mathrm{1}}{\mathrm{2}}\right)\right). \\ $$$$\mathrm{tan}\:\varphi=\mathrm{2}\left({q}−\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{2}{q}−\mathrm{1} \\ $$$${h}={q}+{R}\:\mathrm{sin}\:\varphi={q}+\frac{\left(\mathrm{2}{q}−\mathrm{1}\right){R}}{\:\sqrt{\mathrm{1}+\left(\mathrm{2}{q}−\mathrm{1}\right)^{\mathrm{2}} }} \\ $$$${k}=\frac{\mathrm{25}}{\mathrm{36}}+\left({q}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +{R}\:\mathrm{cos}\:\varphi=\frac{\mathrm{25}}{\mathrm{36}}+\left({q}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{{R}}{\:\sqrt{\mathrm{1}+\left(\mathrm{2}{q}−\mathrm{1}\right)^{\mathrm{2}} }} \\ $$$${q}+\frac{\left(\mathrm{2}{q}−\mathrm{1}\right){R}}{\:\sqrt{\mathrm{1}+\left(\mathrm{2}{q}−\mathrm{1}\right)^{\mathrm{2}} }}={p}+\frac{\mathrm{2}{pR}}{\:\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }} \\ $$$$\frac{\mathrm{25}}{\mathrm{36}}+\left({q}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{{R}}{\:\sqrt{\mathrm{1}+\left(\mathrm{2}{q}−\mathrm{1}\right)^{\mathrm{2}} }}=\frac{\mathrm{1}}{\mathrm{4}}−{p}^{\mathrm{2}} +\frac{{R}}{\:\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }} \\ $$
Commented by ajfour last updated on 09/Nov/25
Thank you sir, i ll sure look into.
$${Thank}\:{you}\:{sir},\:{i}\:{ll}\:{sure}\:{look}\:{into}. \\ $$

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