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# Question-131420

Question Number 131420 by ajfour last updated on 04/Feb/21
Commented by ajfour last updated on 04/Feb/21
$${In}\:{terms}\:{of}\:{ellipse}\:{parameters} \\$$$${a}\:{and}\:{b},\:{find}\:{radius}\:{of}\:{the} \\$$$${equal}\:{radii}\:{circles}. \\$$
Answered by mr W last updated on 04/Feb/21
Commented by mr W last updated on 05/Feb/21
$${let}\:\mu=\frac{{b}}{{a}} \\$$$${say}\:{P}\left({a}+{a}\:\mathrm{cos}\:\phi,\:{b}+{b}\:\mathrm{sin}\:\phi\right) \\$$$$\mathrm{tan}\:\theta=\frac{{b}}{{a}\:\mathrm{tan}\:\phi}=\frac{\mu}{\mathrm{tan}\:\phi} \\$$$$\mathrm{tan}\:\phi=\frac{\mu}{\mathrm{tan}\:\theta} \\$$$$\mathrm{sin}\:\phi=\frac{\mu}{\:\sqrt{\mu^{\mathrm{2}} +\mathrm{tan}^{\mathrm{2}} \:\theta}} \\$$$$\mathrm{cos}\:\phi=\frac{\mathrm{tan}\:\theta}{\:\sqrt{\mu^{\mathrm{2}} +\mathrm{tan}^{\mathrm{2}} \:\theta}} \\$$$${y}={b}+{b}\:\mathrm{sin}\:\phi−\frac{\mu}{\mathrm{tan}\:\phi}\:\left({x}−{a}−{a}\:\mathrm{cos}\:\phi\right) \\$$$$\mathrm{0}={b}\left(\mathrm{1}+\mathrm{sin}\:\phi\right)−\frac{\mu}{\mathrm{tan}\:\phi}\:\left({x}_{{A}} −{a}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\right) \\$$$$\left(\mathrm{1}+\mathrm{sin}\:\phi\right)=\frac{\mathrm{1}}{\mathrm{tan}\:\phi}\:\left(\frac{{x}_{{A}} }{{a}}−\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\right) \\$$$$\Rightarrow\frac{{x}_{{A}} }{{a}}=\mathrm{1}+\mathrm{cos}\:\phi+\left(\mathrm{1}+\mathrm{sin}\:\phi\right)\mathrm{tan}\:\phi \\$$$$\Rightarrow\frac{{x}_{{A}} }{{a}}=\mathrm{1}+\frac{\mathrm{tan}\:\theta}{\:\sqrt{\mu^{\mathrm{2}} +\mathrm{tan}^{\mathrm{2}} \:\theta}}+\left(\mathrm{1}+\frac{\mu}{\:\sqrt{\mu^{\mathrm{2}} +\mathrm{tan}^{\mathrm{2}} \:\theta}}\right)\frac{\mu}{\mathrm{tan}\:\theta} \\$$$$\Rightarrow\frac{{x}_{{A}} }{{a}}=\mathrm{1}+\frac{\mu}{\mathrm{tan}\:\theta}+\sqrt{\mathrm{1}+\left(\frac{\mu}{\mathrm{tan}\:\theta}\right)^{\mathrm{2}} } \\$$$$\\$$$${say}\:{Q}\left({a}+{a}\:\mathrm{cos}\:\alpha,\:{b}−{b}\:\mathrm{sin}\:\alpha\right) \\$$$$\mathrm{tan}\:\varphi=\frac{\mu}{\mathrm{tan}\:\alpha} \\$$$$\mathrm{tan}\:\alpha=\frac{\mu}{\mathrm{tan}\:\varphi} \\$$$$\mathrm{sin}\:\alpha=\frac{\mu}{\:\sqrt{\mu^{\mathrm{2}} +\mathrm{tan}^{\mathrm{2}} \:\varphi}} \\$$$$\mathrm{cos}\:\alpha=\frac{\mathrm{tan}\:\varphi}{\:\sqrt{\mu^{\mathrm{2}} +\mathrm{tan}^{\mathrm{2}} \:\varphi}} \\$$$${x}_{{C}} ={a}+{a}\:\mathrm{cos}\:\alpha+{r}\:\mathrm{sin}\:\varphi={x}_{{A}} −\frac{{r}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}} \\$$$$\mathrm{1}+\mathrm{cos}\:\alpha+\lambda\:\mathrm{sin}\:\varphi=\mathrm{1}+\frac{\mu}{\mathrm{tan}\:\theta}+\sqrt{\mathrm{1}+\left(\frac{\mu}{\mathrm{tan}\:\theta}\right)^{\mathrm{2}} }−\frac{\lambda}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}} \\$$$${let}\:\lambda=\frac{{r}}{{a}} \\$$$$\frac{\mathrm{tan}\:\varphi}{\:\sqrt{\mu^{\mathrm{2}} +\mathrm{tan}^{\mathrm{2}} \:\varphi}}+\lambda\left(\mathrm{sin}\:\varphi+\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}\right)=\frac{\mu}{\mathrm{tan}\:\theta}+\sqrt{\mathrm{1}+\left(\frac{\mu}{\mathrm{tan}\:\theta}\right)^{\mathrm{2}} } \\$$$${y}_{{C}} ={b}−{b}\:\mathrm{sin}\:\alpha−{r}\:\mathrm{cos}\:\varphi={r} \\$$$$\mu\left(\mathrm{1}−\mathrm{sin}\:\alpha\right)=\lambda\left(\mathrm{1}+\mathrm{cos}\:\varphi\right) \\$$$$\lambda=\frac{{r}}{{a}}=\frac{\mu}{\mathrm{1}+\mathrm{cos}\:\varphi}\left(\mathrm{1}−\frac{\mu}{\:\sqrt{\mu^{\mathrm{2}} +\mathrm{tan}^{\mathrm{2}} \:\varphi}}\right) \\$$$$\frac{\mathrm{tan}\:\varphi}{\:\sqrt{\mu^{\mathrm{2}} +\mathrm{tan}^{\mathrm{2}} \:\varphi}}+\frac{\mu}{\mathrm{1}+\mathrm{cos}\:\varphi}\left(\mathrm{1}−\frac{\mu}{\:\sqrt{\mu^{\mathrm{2}} +\mathrm{tan}^{\mathrm{2}} \:\varphi}}\right)\left(\mathrm{sin}\:\varphi+\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}\right)=\frac{\mu}{\mathrm{tan}\:\theta}+\sqrt{\mathrm{1}+\left(\frac{\mu}{\mathrm{tan}\:\theta}\right)^{\mathrm{2}} }\:\:\:…\left({I}\right) \\$$$${similarly} \\$$$$\frac{{r}}{{b}}=\frac{\mathrm{1}}{\mu\left(\mathrm{1}+\mathrm{cos}\:\delta\right)}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mu^{\mathrm{2}} \:\mathrm{tan}^{\mathrm{2}} \:\delta}}\right) \\$$$$\frac{\mu\:\mathrm{tan}\:\delta}{\:\sqrt{\mathrm{1}+\mu^{\mathrm{2}} \:\mathrm{tan}^{\mathrm{2}} \:\delta}}+\frac{\mathrm{1}}{\mu\left(\mathrm{1}+\mathrm{cos}\:\delta\right)}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mu^{\mathrm{2}} \:\mathrm{tan}^{\mathrm{2}} \:\delta}}\right)\left(\mathrm{sin}\:\delta+\frac{\mathrm{1}}{\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}\right)}\right)=\frac{\mathrm{tan}\:\theta}{\mu}+\sqrt{\mathrm{1}+\left(\frac{\mathrm{tan}\:\theta}{\mu}\right)^{\mathrm{2}} }\:\:\:…\left({II}\right) \\$$$$\frac{\mathrm{1}}{\mu\left(\mathrm{1}+\mathrm{cos}\:\delta\right)}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mu^{\mathrm{2}} \:\mathrm{tan}^{\mathrm{2}} \:\delta}}\right)=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{cos}\:\varphi}\left(\mathrm{1}−\frac{\mu}{\:\sqrt{\mu^{\mathrm{2}} +\mathrm{tan}^{\mathrm{2}} \:\varphi}}\right)\:\:\:…\left({III}\right) \\$$$${three}\:{eqn}.\:{for}\:{three}\:{unknowns}:\:\theta,\varphi,\delta. \\$$$$…… \\$$
Answered by ajfour last updated on 04/Feb/21
Commented by mr W last updated on 05/Feb/21
$${i}\:{found}\:{no}\:{way}\:{to}\:{reduce}\:{the}\:{problem} \\$$$${to}\:{two}\:{equations}\:{with}\:{two}\:{unknowns}. \\$$