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Question Number 173677 by naka3546 last updated on 16/Jul/22

Find  tangent  equation  of    two  circles :  L_1  : (x−5)^2  + (y+1)^2  = 9  L_2  : (x+4)^2  + (y−11)^2  = 144

$$\mathrm{Find}\:\:\mathrm{tangent}\:\:\mathrm{equation}\:\:\mathrm{of}\:\: \\ $$$$\mathrm{two}\:\:\mathrm{circles}\:: \\ $$$${L}_{\mathrm{1}} \::\:\left({x}−\mathrm{5}\right)^{\mathrm{2}} \:+\:\left({y}+\mathrm{1}\right)^{\mathrm{2}} \:=\:\mathrm{9} \\ $$$${L}_{\mathrm{2}} \::\:\left({x}+\mathrm{4}\right)^{\mathrm{2}} \:+\:\left({y}−\mathrm{11}\right)^{\mathrm{2}} \:=\:\mathrm{144} \\ $$

Commented by kaivan.ahmadi last updated on 16/Jul/22

L_1 : x^2 +y^2 −10x+2y=−17  L_2 : x^2 +y^2 +8x−22y=7  L_1 −L_2 : −18x+24y=−24  y=(3/4)x−1

$${L}_{\mathrm{1}} :\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{2}{y}=−\mathrm{17} \\ $$$${L}_{\mathrm{2}} :\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{8}{x}−\mathrm{22}{y}=\mathrm{7} \\ $$$${L}_{\mathrm{1}} −{L}_{\mathrm{2}} :\:−\mathrm{18}{x}+\mathrm{24}{y}=−\mathrm{24} \\ $$$${y}=\frac{\mathrm{3}}{\mathrm{4}}{x}−\mathrm{1} \\ $$

Commented by naka3546 last updated on 16/Jul/22

Thank  you, sir.

$$\mathrm{Thank}\:\:\mathrm{you},\:\mathrm{sir}. \\ $$

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