Question Number 145575 by mnjuly1970 last updated on 06/Jul/21
Answered by ajfour last updated on 06/Jul/21
Commented by ajfour last updated on 06/Jul/21
$${Let}\:{the}\:{intersection}\:{be}\:{the} \\ $$$${origin}. \\ $$$${eq}.\:{of}\:{circle} \\ $$$$\:\:\left({x}−\mathrm{2}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{36} \\ $$$$\:{y}=\pm\sqrt{\mathrm{3}}{x} \\ $$$$\Rightarrow\:\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{32}=\mathrm{0} \\ $$$$\:\:\:{x}^{\mathrm{2}} −{x}−\mathrm{8}=\mathrm{0} \\ $$$$\:\:\:{x}_{{A}} ,\:{x}_{{B}} =\frac{\mathrm{1}\pm\sqrt{\mathrm{33}}}{\mathrm{2}} \\ $$$${AB}^{\mathrm{2}} =\left({x}_{{A}} −{x}_{{B}} \right)^{\mathrm{2}} +\left({y}_{{A}} −{y}_{{B}} \right)^{\mathrm{2}} \\ $$$$\:\:=\left({x}_{{A}} −{x}_{{B}} \right)^{\mathrm{2}} +\mathrm{3}\left({x}_{{A}} +{x}_{{B}} \right)^{\mathrm{2}} \\ $$$$\:\:=\mathrm{33}+\mathrm{3}=\mathrm{36} \\ $$$${AB}=\mathrm{6}\:\:\left(={radius}\right) \\ $$
Commented by ajfour last updated on 06/Jul/21
please understand the mistake in labeling the
equation of the red and blue lines.
Commented by mnjuly1970 last updated on 06/Jul/21
$$\:{grateful}\:{mr}\:{ajfor}\:… \\ $$
Answered by mr W last updated on 06/Jul/21
Commented by mr W last updated on 06/Jul/21
$$\left(\frac{\sqrt{\mathrm{3}}{d}}{\mathrm{2}}\right)^{\mathrm{2}} =\left({a}+\frac{{d}}{\mathrm{2}}\right)\left({b}−\frac{{d}}{\mathrm{2}}\right) \\ $$$$\mathrm{2}{d}^{\mathrm{2}} −\left({b}−{a}\right){d}−\mathrm{2}{ab}=\mathrm{0} \\ $$$${d}=\frac{\left({b}−{a}\right)+\sqrt{\left({b}−{a}\right)^{\mathrm{2}} +\mathrm{16}{ab}}}{\mathrm{4}} \\ $$$${similarly} \\ $$$${c}=\frac{−\left({b}−{a}\right)+\sqrt{\left({b}−{a}\right)^{\mathrm{2}} +\mathrm{16}{ab}}}{\mathrm{4}} \\ $$$${x}^{\mathrm{2}} ={c}^{\mathrm{2}} +{d}^{\mathrm{2}} −\mathrm{2}{cd}\mathrm{cos}\:\mathrm{60}° \\ $$$$\:\:\:\:\:=\left({c}+{d}\right)^{\mathrm{2}} −\mathrm{3}{cd} \\ $$$$\:\:\:\:\:=\frac{\left({b}−{a}\right)^{\mathrm{2}} +\mathrm{16}{ab}}{\mathrm{4}}−\mathrm{3}{ab} \\ $$$$\:\:\:\:\:=\left(\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{x}=\frac{{a}+{b}}{\mathrm{2}}={radius}=\frac{\mathrm{4}+\mathrm{8}}{\mathrm{2}}=\mathrm{6} \\ $$
Commented by mnjuly1970 last updated on 07/Jul/21
$$\:{excellent},''\:{mr}\:\mathrm{W}''\:{as}\:{always}. \\ $$