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Find-the-equation-of-the-circle-circumscribing-the-triangle-form-by-lines-3x-y-5-0-x-y-1-0-and-2x-y-4-0-

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Question Number 6192 by sanusihammed last updated on 17/Jun/16
Find the equation of the circle circumscribing the triangle form by lines 3x + y − 5 = 0, x + y + 1 = 0, and 2x + y − 4 = 0
$${Find}\:{the}\:{equation}\:{of}\:{the}\:{circle}\:{circumscribing}\:{the}\:{triangle}\:{form}\:{by}\: \\ $$$${lines}\:\:\mathrm{3}{x}\:+\:{y}\:−\:\mathrm{5}\:=\:\mathrm{0},\:{x}\:+\:{y}\:+\:\mathrm{1}\:=\:\mathrm{0},\:{and}\:\mathrm{2}{x}\:+\:{y}\:−\:\mathrm{4}\:=\:\mathrm{0} \\ $$
Answered by Rasheed Soomro last updated on 18/Jun/16
3x + y − 5 = 0......(i) x + y + 1 = 0........(ii) 2x + y − 4 = 0.......(iii) (i)−(ii) : 2x−6=0⇒x=3⇒y=−4 : (3,−4) (iii)−(ii): x−5=0⇒x=5⇒y=−6: (5,−6) (i)−(iii): x−1=0⇒x=1⇒y=2: (1,2) Let equation passing through vertices (3,−4),(5,−6) and (1,2) is: x^2 +y^2 +2gx+2fy+c=0 Since (3,−4),(5,−6) and (1,2) lie on the circle, they′ll satisfy the equagion (3)^2 +(−4)^2 +2g(3)+2f(−4)+c=0 6g−8f+c=−25................I (5)^2 +(−6)^2 +2g(5)+2f(−6)+c=0 10g−12f+c=−61...........II (1)^2 +(2)^2 +2g(1)+2f(2)+c=0 2g+4f+c=−5...............III II−I: 4g−4f=−36⇒g−f=−9...........A II−III: 8g−16f=−56⇒g−2f=−7......B A−B: f=−2, A: g−(−2)= −9⇒g=−11 III: 2(−11)+4(−2)+c=−5⇒c=25 The required equation: x^2 +y^2 +2gx+2fy+c=0 ⇒ x^2 +y^2 +2(−11)x+2(−2)y+(25)=0 ⇒x^2 +y^2 −22x−4y+25=0
$$\mathrm{3}{x}\:+\:{y}\:−\:\mathrm{5}\:=\:\mathrm{0}……\left({i}\right) \\ $$$${x}\:+\:{y}\:+\:\mathrm{1}\:=\:\mathrm{0}……..\left({ii}\right) \\ $$$$\mathrm{2}{x}\:+\:{y}\:−\:\mathrm{4}\:=\:\mathrm{0}…….\left({iii}\right) \\ $$$$\left({i}\right)−\left({ii}\right)\::\:\:\mathrm{2}{x}−\mathrm{6}=\mathrm{0}\Rightarrow{x}=\mathrm{3}\Rightarrow{y}=−\mathrm{4}\::\:\left(\mathrm{3},−\mathrm{4}\right) \\ $$$$\left({iii}\right)−\left({ii}\right):\:{x}−\mathrm{5}=\mathrm{0}\Rightarrow{x}=\mathrm{5}\Rightarrow{y}=−\mathrm{6}:\:\left(\mathrm{5},−\mathrm{6}\right) \\ $$$$\left({i}\right)−\left({iii}\right):\:{x}−\mathrm{1}=\mathrm{0}\Rightarrow{x}=\mathrm{1}\Rightarrow{y}=\mathrm{2}:\:\left(\mathrm{1},\mathrm{2}\right) \\ $$$$ \\ $$$${Let}\:{equation}\:{passing}\:{through}\:{vertices} \\ $$$$\left(\mathrm{3},−\mathrm{4}\right),\left(\mathrm{5},−\mathrm{6}\right)\:{and}\:\left(\mathrm{1},\mathrm{2}\right)\:{is}: \\ $$$$\:\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{gx}+\mathrm{2}{fy}+{c}=\mathrm{0} \\ $$$${Since}\:\left(\mathrm{3},−\mathrm{4}\right),\left(\mathrm{5},−\mathrm{6}\right)\:{and}\:\left(\mathrm{1},\mathrm{2}\right)\:{lie}\:{on}\:{the}\: \\ $$$${circle},\:{they}'{ll}\:{satisfy}\:{the}\:{equagion} \\ $$$$\:\:\:\:\:\:\:\:\:\:\left(\mathrm{3}\right)^{\mathrm{2}} +\left(−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{2}{g}\left(\mathrm{3}\right)+\mathrm{2}{f}\left(−\mathrm{4}\right)+{c}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{6}{g}−\mathrm{8}{f}+{c}=−\mathrm{25}…………….{I} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{5}\right)^{\mathrm{2}} +\left(−\mathrm{6}\right)^{\mathrm{2}} +\mathrm{2}{g}\left(\mathrm{5}\right)+\mathrm{2}{f}\left(−\mathrm{6}\right)+{c}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{10}{g}−\mathrm{12}{f}+{c}=−\mathrm{61}………..{II} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{2}\right)^{\mathrm{2}} +\mathrm{2}{g}\left(\mathrm{1}\right)+\mathrm{2}{f}\left(\mathrm{2}\right)+{c}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}{g}+\mathrm{4}{f}+{c}=−\mathrm{5}……………{III} \\ $$$$ \\ $$$${II}−{I}:\:\mathrm{4}{g}−\mathrm{4}{f}=−\mathrm{36}\Rightarrow{g}−{f}=−\mathrm{9}………..{A} \\ $$$${II}−{III}:\:\mathrm{8}{g}−\mathrm{16}{f}=−\mathrm{56}\Rightarrow{g}−\mathrm{2}{f}=−\mathrm{7}……{B} \\ $$$${A}−{B}:\:{f}=−\mathrm{2},\:\:{A}:\:{g}−\left(−\mathrm{2}\right)=\:−\mathrm{9}\Rightarrow{g}=−\mathrm{11} \\ $$$${III}:\:\mathrm{2}\left(−\mathrm{11}\right)+\mathrm{4}\left(−\mathrm{2}\right)+{c}=−\mathrm{5}\Rightarrow{c}=\mathrm{25} \\ $$$${The}\:{required}\:{equation}: \\ $$$$\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{gx}+\mathrm{2}{fy}+{c}=\mathrm{0} \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}\left(−\mathrm{11}\right){x}+\mathrm{2}\left(−\mathrm{2}\right){y}+\left(\mathrm{25}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{22}{x}−\mathrm{4}{y}+\mathrm{25}=\mathrm{0} \\ $$
Commented by sanusihammed last updated on 18/Jun/16
Thanks so much
$${Thanks}\:{so}\:{much} \\ $$
Commented by sanusihammed last updated on 18/Jun/16
i really appreciate
$${i}\:{really}\:{appreciate} \\ $$

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