Question Number 45690 by ajfour last updated on 15/Oct/18
Commented by ajfour last updated on 15/Oct/18
$${Radius}\:{of}\:{circle}\:{is}\:{unity}. \\ $$$${Find}\:{b}\:{in}\:{terms}\:{of}\:{a}. \\ $$
Answered by MrW3 last updated on 15/Oct/18
Commented by ajfour last updated on 15/Oct/18
$${Thank}\:{you}\:{so}\:{much}\:{Sir};\:{it}\:{is} \\ $$$${indeed}\:{a}\:{matter}\:{of}\:{faith}. \\ $$$$\mathcal{BEAUTIFUL}\:! \\ $$
Commented by MrW3 last updated on 15/Oct/18
$${thank}\:{you}\:{too}\:{sir}! \\ $$
Commented by MrW3 last updated on 15/Oct/18
$$\mathrm{tan}\:\alpha=\frac{\sqrt{\left(\mathrm{2}{R}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{a}}=\frac{\sqrt{\mathrm{4}{R}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{a}} \\ $$$$\gamma=\mathrm{2}\alpha \\ $$$${PT}={R}\:\mathrm{tan}\:\gamma={R}\:\mathrm{tan}\:\left(\mathrm{2}\alpha\right)={R}\frac{\mathrm{2}\sqrt{\mathrm{4}{R}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{a}\left(\mathrm{1}−\frac{\mathrm{4}{R}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)}=\frac{{aR}\sqrt{\mathrm{4}{R}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{a}^{\mathrm{2}} −\mathrm{2}{R}^{\mathrm{2}} } \\ $$$$\mathrm{tan}\:\beta=\frac{{PT}}{\mathrm{2}{R}}=\frac{{a}\sqrt{\mathrm{4}{R}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{\mathrm{2}\left({a}^{\mathrm{2}} −\mathrm{2}{R}^{\mathrm{2}} \right)} \\ $$$$\mathrm{cos}\:\beta=\frac{\mathrm{2}\left({a}^{\mathrm{2}} −\mathrm{2}{R}^{\mathrm{2}} \right)}{\:\sqrt{{a}^{\mathrm{2}} \left(\mathrm{4}{R}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)+\mathrm{4}\left({a}^{\mathrm{2}} −\mathrm{2}{R}^{\mathrm{2}} \right)^{\mathrm{2}} }} \\ $$$$\mathrm{cos}\:\beta=\frac{\mathrm{2}\left({a}^{\mathrm{2}} −\mathrm{2}{R}^{\mathrm{2}} \right)}{\:\sqrt{\mathrm{3}{a}^{\mathrm{4}} +\mathrm{16}{R}^{\mathrm{4}} −\mathrm{12}{a}^{\mathrm{2}} {R}^{\mathrm{2}} }} \\ $$$$\mathrm{cos}\:\beta=\frac{\mathrm{2}\left({a}^{\mathrm{2}} −\mathrm{2}{R}^{\mathrm{2}} \right)}{\:\sqrt{\mathrm{3}\left({a}^{\mathrm{2}} −\mathrm{2}{R}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{4}{R}^{\mathrm{4}} }} \\ $$$${b}=\mathrm{2}{R}\:\mathrm{cos}\:\beta \\ $$$$\Rightarrow{b}=\frac{\mathrm{4}{R}\left({a}^{\mathrm{2}} −\mathrm{2}{R}^{\mathrm{2}} \right)}{\:\sqrt{\mathrm{3}\left({a}^{\mathrm{2}} −\mathrm{2}{R}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{4}{R}^{\mathrm{4}} }} \\ $$$${for}\:{R}=\mathrm{1}: \\ $$$$\Rightarrow{b}=\frac{\mathrm{4}\left({a}^{\mathrm{2}} −\mathrm{2}\right)}{\:\sqrt{\mathrm{3}\left({a}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} +\mathrm{4}}} \\ $$