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Question Number 29627 by ajfour last updated on 10/Feb/18

Commented by ajfour last updated on 10/Feb/18

Also find (a/R) if length of AB is  equal to a .

$${Also}\:{find}\:\frac{\boldsymbol{{a}}}{\boldsymbol{{R}}}\:{if}\:{length}\:{of}\:{AB}\:{is} \\ $$$${equal}\:{to}\:\boldsymbol{{a}}\:. \\ $$

Answered by mrW2 last updated on 10/Feb/18

O=center of circle  OP=(√(a^2 +R^2 ))  PT=(√(a^2 +R^2 ))+R  ((BT)/(PT))=((OQ)/(QT))  BT=((R((√(a^2 +R^2 ))+R))/a)  AB^2 =AT^2 +BT^2 =4R^2 +((R^2 (2R^2 +a^2 +2R(√(a^2 +R^2 ))))/a^2 )  =((R^2 (2R^2 +5a^2 +2R(√(a^2 +R^2 ))))/a^2 )  ⇒AB=((R(√(2R^2 +5a^2 +2R(√(a^2 +R^2 )))))/a)    if AB=a  ((R^2 (2R^2 +5a^2 +2R(√(a^2 +R^2 ))))/a^2 )=a^2   ⇒2+5((a/R))^2 +2(√(1+((a/R))^2 ))=((a/R))^4   ⇒(a/R)≈2.4875 (via graph)

$${O}={center}\:{of}\:{circle} \\ $$$${OP}=\sqrt{{a}^{\mathrm{2}} +{R}^{\mathrm{2}} } \\ $$$${PT}=\sqrt{{a}^{\mathrm{2}} +{R}^{\mathrm{2}} }+{R} \\ $$$$\frac{{BT}}{{PT}}=\frac{{OQ}}{{QT}} \\ $$$${BT}=\frac{{R}\left(\sqrt{{a}^{\mathrm{2}} +{R}^{\mathrm{2}} }+{R}\right)}{{a}} \\ $$$${AB}^{\mathrm{2}} ={AT}^{\mathrm{2}} +{BT}^{\mathrm{2}} =\mathrm{4}{R}^{\mathrm{2}} +\frac{{R}^{\mathrm{2}} \left(\mathrm{2}{R}^{\mathrm{2}} +{a}^{\mathrm{2}} +\mathrm{2}{R}\sqrt{{a}^{\mathrm{2}} +{R}^{\mathrm{2}} }\right)}{{a}^{\mathrm{2}} } \\ $$$$=\frac{{R}^{\mathrm{2}} \left(\mathrm{2}{R}^{\mathrm{2}} +\mathrm{5}{a}^{\mathrm{2}} +\mathrm{2}{R}\sqrt{{a}^{\mathrm{2}} +{R}^{\mathrm{2}} }\right)}{{a}^{\mathrm{2}} } \\ $$$$\Rightarrow{AB}=\frac{{R}\sqrt{\mathrm{2}{R}^{\mathrm{2}} +\mathrm{5}{a}^{\mathrm{2}} +\mathrm{2}{R}\sqrt{{a}^{\mathrm{2}} +{R}^{\mathrm{2}} }}}{{a}} \\ $$$$ \\ $$$${if}\:{AB}={a} \\ $$$$\frac{{R}^{\mathrm{2}} \left(\mathrm{2}{R}^{\mathrm{2}} +\mathrm{5}{a}^{\mathrm{2}} +\mathrm{2}{R}\sqrt{{a}^{\mathrm{2}} +{R}^{\mathrm{2}} }\right)}{{a}^{\mathrm{2}} }={a}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}+\mathrm{5}\left(\frac{{a}}{{R}}\right)^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{1}+\left(\frac{{a}}{{R}}\right)^{\mathrm{2}} }=\left(\frac{{a}}{{R}}\right)^{\mathrm{4}} \\ $$$$\Rightarrow\frac{{a}}{{R}}\approx\mathrm{2}.\mathrm{4875}\:\left({via}\:{graph}\right) \\ $$

Commented by ajfour last updated on 10/Feb/18

Thank you Sir.

$${Thank}\:{you}\:{Sir}. \\ $$

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