Question Number 199149 by ajfour last updated on 28/Oct/23
Commented by ajfour last updated on 28/Oct/23
$${No};\:\:{rather}\:{a}=\mathrm{1}<{b}.\:\:{Find}\:{R}={f}\left({b}\right). \\ $$
Answered by ajfour last updated on 28/Oct/23
Answered by mr W last updated on 29/Oct/23
Commented by mr W last updated on 29/Oct/23
![(R/(cos θ))−(a/(tan ((π/4)−(θ/2))))=(√((R+a)^2 −a^2 )) let α=(a/R), β=(b/R) (1/(cos θ))−(α/(tan ((π/4)−(θ/2))))=(√(1+2α)) (2α+1)+2 tan ((π/4)−(θ/2))(√(2α+1))−(1+((2 tan ((π/4)−(θ/2)))/(cos θ)))=0 (√(2α+1))=(√(1+tan^2 ((π/4)−(θ/2))+((2 tan ((π/4)−(θ/2)))/(cos θ))))−tan ((π/4)−(θ/2)) ⇒α(θ)=(1/2)[(√(1+tan^2 ((π/4)−(θ/2))+((2 tan ((π/4)−(θ/2)))/(cos θ))))−tan ((π/4)−(θ/2))]^2 −(1/2) similarly ⇒β(θ)=(1/2)[(√(1+tan^2 (θ/2)+((2 tan (θ/2))/(sin θ))))−tan (θ/2)]^2 −(1/2) from ((β(θ))/(a(θ)))=(b/a) we solve for θ and then get R=(b/(β(θ))). example: a=1, b=3 ⇒θ≈0.5658 ⇒R≈15.1054](https://www.tinkutara.com/question/Q199179.png)
$$\frac{{R}}{\mathrm{cos}\:\theta}−\frac{{a}}{\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}\right)}=\sqrt{\left({R}+{a}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$${let}\:\alpha=\frac{{a}}{{R}},\:\beta=\frac{{b}}{{R}} \\ $$$$\frac{\mathrm{1}}{\mathrm{cos}\:\theta}−\frac{\alpha}{\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}\right)}=\sqrt{\mathrm{1}+\mathrm{2}\alpha} \\ $$$$\left(\mathrm{2}\alpha+\mathrm{1}\right)+\mathrm{2}\:\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}\right)\sqrt{\mathrm{2}\alpha+\mathrm{1}}−\left(\mathrm{1}+\frac{\mathrm{2}\:\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}\right)}{\mathrm{cos}\:\theta}\right)=\mathrm{0} \\ $$$$\sqrt{\mathrm{2}\alpha+\mathrm{1}}=\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\left(\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}\right)+\frac{\mathrm{2}\:\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}\right)}{\mathrm{cos}\:\theta}}−\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}\right) \\ $$$$\Rightarrow\alpha\left(\theta\right)=\frac{\mathrm{1}}{\mathrm{2}}\left[\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\left(\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}\right)+\frac{\mathrm{2}\:\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}\right)}{\mathrm{cos}\:\theta}}−\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}\right)\right]^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${similarly} \\ $$$$\Rightarrow\beta\left(\theta\right)=\frac{\mathrm{1}}{\mathrm{2}}\left[\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}+\frac{\mathrm{2}\:\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}{\mathrm{sin}\:\theta}}−\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\right]^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${from}\:\frac{\beta\left(\theta\right)}{{a}\left(\theta\right)}=\frac{{b}}{{a}}\:{we}\:{solve}\:{for}\:\theta\:{and} \\ $$$${then}\:{get}\:{R}=\frac{{b}}{\beta\left(\theta\right)}. \\ $$$$ \\ $$$${example}: \\ $$$${a}=\mathrm{1},\:{b}=\mathrm{3} \\ $$$$\Rightarrow\theta\approx\mathrm{0}.\mathrm{5658}\:\Rightarrow{R}\approx\mathrm{15}.\mathrm{1054} \\ $$
Commented by mr W last updated on 29/Oct/23
Commented by ajfour last updated on 29/Oct/23
$${Thank}\:{you}\:{sir}.\:{I}\:{am}\:{trying}\:{still}\:{for} \\ $$$${an}\:{exact}\:{one}. \\ $$
Answered by mr W last updated on 29/Oct/23
Commented by mr W last updated on 29/Oct/23
![equ. of tangent line: x cos θ+y sin θ−R=0 center of circle A((√(R(R+2a))), a) center of circle B(b, (√(R(R+2b))) (√(R(R+2a))) cos θ+a sin θ−R=−a b cos θ+(√(R(R+2a))) sin θ−R=−b ⇒sin θ=(((R−b)(√(R(R+2a)))−(R−a)b)/( R(√((R+2a)(R+2b)))−ab)) ⇒cos θ=(((R−a)(√(R(R+2b)))−(R−b)a)/( R(√((R+2a)(R+2b)))−ab)) [(R−b)(√(R(R+2a)))−(R−a)b]^2 +[(R−a)(√(R(R+2b)))−(R−b)a]^2 =[ R(√((R+2a)(R+2b)))−ab]^2 example: (b/a)=3 ⇒(R/a)≈15.104081062050845](https://www.tinkutara.com/question/Q199224.png)
$${equ}.\:{of}\:{tangent}\:{line}: \\ $$$${x}\:\mathrm{cos}\:\theta+{y}\:\mathrm{sin}\:\theta−{R}=\mathrm{0} \\ $$$${center}\:{of}\:{circle}\:{A}\left(\sqrt{{R}\left({R}+\mathrm{2}{a}\right)},\:{a}\right) \\ $$$${center}\:{of}\:{circle}\:{B}\left({b},\:\sqrt{{R}\left({R}+\mathrm{2}{b}\right.}\right) \\ $$$$\sqrt{{R}\left({R}+\mathrm{2}{a}\right)}\:\mathrm{cos}\:\theta+{a}\:\mathrm{sin}\:\theta−{R}=−{a} \\ $$$${b}\:\mathrm{cos}\:\theta+\sqrt{{R}\left({R}+\mathrm{2}{a}\right)}\:\mathrm{sin}\:\theta−{R}=−{b} \\ $$$$\Rightarrow\mathrm{sin}\:\theta=\frac{\left({R}−{b}\right)\sqrt{{R}\left({R}+\mathrm{2}{a}\right)}−\left({R}−{a}\right){b}}{\:{R}\sqrt{\left({R}+\mathrm{2}{a}\right)\left({R}+\mathrm{2}{b}\right)}−{ab}} \\ $$$$\Rightarrow\mathrm{cos}\:\theta=\frac{\left({R}−{a}\right)\sqrt{{R}\left({R}+\mathrm{2}{b}\right)}−\left({R}−{b}\right){a}}{\:{R}\sqrt{\left({R}+\mathrm{2}{a}\right)\left({R}+\mathrm{2}{b}\right)}−{ab}} \\ $$$$\left[\left({R}−{b}\right)\sqrt{{R}\left({R}+\mathrm{2}{a}\right)}−\left({R}−{a}\right){b}\right]^{\mathrm{2}} +\left[\left({R}−{a}\right)\sqrt{{R}\left({R}+\mathrm{2}{b}\right)}−\left({R}−{b}\right){a}\right]^{\mathrm{2}} =\left[\:{R}\sqrt{\left({R}+\mathrm{2}{a}\right)\left({R}+\mathrm{2}{b}\right)}−{ab}\right]^{\mathrm{2}} \\ $$$$ \\ $$$${example}: \\ $$$$\frac{{b}}{{a}}=\mathrm{3}\:\Rightarrow\frac{{R}}{{a}}\approx\mathrm{15}.\mathrm{104081062050845} \\ $$
Commented by ajfour last updated on 29/Oct/23
$${This}\:{is}\:{what}\:{i}\:{had}\:{hoped}!\:{thanks}\:{Sir}. \\ $$