Question Number 29627 by ajfour last updated on 10/Feb/18
Commented by ajfour last updated on 10/Feb/18
$${Also}\:{find}\:\frac{\boldsymbol{{a}}}{\boldsymbol{{R}}}\:{if}\:{length}\:{of}\:{AB}\:{is} \\ $$$${equal}\:{to}\:\boldsymbol{{a}}\:. \\ $$
Answered by mrW2 last updated on 10/Feb/18
$${O}={center}\:{of}\:{circle} \\ $$$${OP}=\sqrt{{a}^{\mathrm{2}} +{R}^{\mathrm{2}} } \\ $$$${PT}=\sqrt{{a}^{\mathrm{2}} +{R}^{\mathrm{2}} }+{R} \\ $$$$\frac{{BT}}{{PT}}=\frac{{OQ}}{{QT}} \\ $$$${BT}=\frac{{R}\left(\sqrt{{a}^{\mathrm{2}} +{R}^{\mathrm{2}} }+{R}\right)}{{a}} \\ $$$${AB}^{\mathrm{2}} ={AT}^{\mathrm{2}} +{BT}^{\mathrm{2}} =\mathrm{4}{R}^{\mathrm{2}} +\frac{{R}^{\mathrm{2}} \left(\mathrm{2}{R}^{\mathrm{2}} +{a}^{\mathrm{2}} +\mathrm{2}{R}\sqrt{{a}^{\mathrm{2}} +{R}^{\mathrm{2}} }\right)}{{a}^{\mathrm{2}} } \\ $$$$=\frac{{R}^{\mathrm{2}} \left(\mathrm{2}{R}^{\mathrm{2}} +\mathrm{5}{a}^{\mathrm{2}} +\mathrm{2}{R}\sqrt{{a}^{\mathrm{2}} +{R}^{\mathrm{2}} }\right)}{{a}^{\mathrm{2}} } \\ $$$$\Rightarrow{AB}=\frac{{R}\sqrt{\mathrm{2}{R}^{\mathrm{2}} +\mathrm{5}{a}^{\mathrm{2}} +\mathrm{2}{R}\sqrt{{a}^{\mathrm{2}} +{R}^{\mathrm{2}} }}}{{a}} \\ $$$$ \\ $$$${if}\:{AB}={a} \\ $$$$\frac{{R}^{\mathrm{2}} \left(\mathrm{2}{R}^{\mathrm{2}} +\mathrm{5}{a}^{\mathrm{2}} +\mathrm{2}{R}\sqrt{{a}^{\mathrm{2}} +{R}^{\mathrm{2}} }\right)}{{a}^{\mathrm{2}} }={a}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}+\mathrm{5}\left(\frac{{a}}{{R}}\right)^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{1}+\left(\frac{{a}}{{R}}\right)^{\mathrm{2}} }=\left(\frac{{a}}{{R}}\right)^{\mathrm{4}} \\ $$$$\Rightarrow\frac{{a}}{{R}}\approx\mathrm{2}.\mathrm{4875}\:\left({via}\:{graph}\right) \\ $$
Commented by ajfour last updated on 10/Feb/18
$${Thank}\:{you}\:{Sir}. \\ $$