Question Number 203062 by ajfour last updated on 09/Jan/24
Answered by ajfour last updated on 09/Jan/24
$$\mathrm{tan}\:\theta=\frac{{R}−\left(\mathrm{1}−{R}\right)}{{x}−{c}}=\frac{\mathrm{1}}{{x}} \\ $$$$\&\:\:{x}^{\mathrm{2}} ={R}^{\mathrm{2}} −\left(\mathrm{1}−{R}\right)^{\mathrm{2}} =\mathrm{2}{R}−\mathrm{1} \\ $$$$\Rightarrow\:\frac{{x}−{c}}{{x}}=\frac{\mathrm{2}{R}−\mathrm{1}}{\mathrm{1}}={x}^{\mathrm{2}} \\ $$$$\frac{\mathrm{2}{x}−{c}}{{c}}=\frac{{R}}{\mathrm{1}−{R}} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} ={c}^{\mathrm{2}} \left(\mathrm{1}+\frac{{R}}{\mathrm{1}−{R}}\right)^{\mathrm{2}} =\mathrm{4}\left(\mathrm{2}{R}−\mathrm{1}\right) \\ $$$$\left(\mathrm{2}{R}−\mathrm{1}\right)\left(\mathrm{2}{R}−\mathrm{2}\right)^{\mathrm{2}} ={c}^{\mathrm{2}} \\ $$$${t}^{\mathrm{2}} \left({t}+\mathrm{1}\right)={c}^{\mathrm{2}} \\ $$