Question and Answers Forum

All Questions      Topic List

Coordinate Geometry Questions

Previous in All Question      Next in All Question      

Previous in Coordinate Geometry      Next in Coordinate Geometry      

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

$${i}\:{really}\:{appreciate} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com