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Question Number 5982 by Kasih last updated on 08/Jun/16
the center of circle in 2x+y−11=0 determine the equation of circle that passing through (−1,3),(7,−1)
thecenterofcirclein2x+y−11=0determinetheequationofcirclethatpassingthrough(−1,3),(7,−1)
Commented by Rasheed Soomro last updated on 08/Jun/16
the center of circle in 2x+y−11=0 determine the equation of circle that passing through (−1,3),(7,−1) −−−−−−−−−−−−−−− 2x+y−11=0⇒y=11−2x Hence moving point of tbe line (x,y)=(x,11−2x) One of these points is center of the required circle and equidistant from all tbe points of circle: Hence, Distance of(−1,3) from (x,11−2x) =Distance of(7,−1) from (x,11−2x) (√((x+1)^2 +(11−2x−3)^2 ))=(√((x−7)^2 +(11−2x+1)^2 )) (x+1)^2 +(−2x+8)^2 =(x−7)^2 +(−2x+12)^2 x^2 +2x+1+4x^2 −32x+64=x^2 −14x+49+4x^2 −48x+144 2x+1−32x+64=−14x+49−48x+144 −30x+65=−62x+193 32x=128⇒x=4 Therefore center of the required circle is (x,11−2x)=(4,11−2×4)=(4,3) Radius will be distance between (−1,3 ) and (4,3) (√((4+1)^2 +(3−3)^2 ))=5 The equation of required circle (x−4)^2 +(y−3)^2 =5^2 =25 Or another form x^2 +y^2 −8x−6y=0
thecenterofcirclein2x+y−11=0determinetheequationofcirclethatpassingthrough(−1,3),(7,−1)−−−−−−−−−−−−−−−2x+y−11=0⇒y=11−2xHencemovingpointoftbeline(x,y)=(x,11−2x)Oneofthesepointsiscenteroftherequiredcircleandequidistantfromalltbepointsofcircle:Hence,Distanceof(−1,3)from(x,11−2x)=Distanceof(7,−1)from(x,11−2x)(x+1)2+(11−2x−3)2=(x−7)2+(11−2x+1)2(x+1)2+(−2x+8)2=(x−7)2+(−2x+12)2x2+2x+1+4x2−32x+64=x2−14x+49+4x2−48x+1442x+1−32x+64=−14x+49−48x+144−30x+65=−62x+19332x=128⇒x=4Thereforecenteroftherequiredcircleis(x,11−2x)=(4,11−2×4)=(4,3)Radiuswillbedistancebetween(−1,3)and(4,3)(4+1)2+(3−3)2=5Theequationofrequiredcircle(x−4)2+(y−3)2=52=25Oranotherformx2+y2−8x−6y=0

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