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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)
$${the}\:{center}\:{of}\:{circle}\:{in}\:\mathrm{2}{x}+{y}−\mathrm{11}=\mathrm{0} \\ $$$${determine}\:{the}\:{equation}\:{of}\:{circle}\:{that} \\ $$$${passing}\:{through}\:\left(−\mathrm{1},\mathrm{3}\right),\left(\mathrm{7},−\mathrm{1}\right) \\ $$
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
$${the}\:{center}\:{of}\:{circle}\:{in}\:\mathrm{2}{x}+{y}−\mathrm{11}=\mathrm{0} \\ $$$${determine}\:{the}\:{equation}\:{of}\:{circle}\:{that} \\ $$$${passing}\:{through}\:\left(−\mathrm{1},\mathrm{3}\right),\left(\mathrm{7},−\mathrm{1}\right) \\ $$$$−−−−−−−−−−−−−−− \\ $$$$\mathrm{2}{x}+{y}−\mathrm{11}=\mathrm{0}\Rightarrow{y}=\mathrm{11}−\mathrm{2}{x} \\ $$$${Hence}\:{moving}\:{point}\:{of}\:{tbe}\:{line} \\ $$$$\left({x},{y}\right)=\left({x},\mathrm{11}−\mathrm{2}{x}\right) \\ $$$${One}\:{of}\:{these}\:\:{points}\:{is}\:{center}\:{of} \\ $$$${the}\:{required}\:{circle}\:{and}\:{equidistant} \\ $$$${from}\:{all}\:{tbe}\:{points}\:{of}\:{circle}: \\ $$$${Hence}, \\ $$$${Distance}\:{of}\left(−\mathrm{1},\mathrm{3}\right)\:{from}\:\left({x},\mathrm{11}−\mathrm{2}{x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={Distance}\:{of}\left(\mathrm{7},−\mathrm{1}\right)\:{from}\:\left({x},\mathrm{11}−\mathrm{2}{x}\right) \\ $$$$\sqrt{\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{11}−\mathrm{2}{x}−\mathrm{3}\right)^{\mathrm{2}} }=\sqrt{\left({x}−\mathrm{7}\right)^{\mathrm{2}} +\left(\mathrm{11}−\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\left(−\mathrm{2}{x}+\mathrm{8}\right)^{\mathrm{2}} =\left({x}−\mathrm{7}\right)^{\mathrm{2}} +\left(−\mathrm{2}{x}+\mathrm{12}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} −\mathrm{32}{x}+\mathrm{64}={x}^{\mathrm{2}} −\mathrm{14}{x}+\mathrm{49}+\mathrm{4}{x}^{\mathrm{2}} −\mathrm{48}{x}+\mathrm{144} \\ $$$$\mathrm{2}{x}+\mathrm{1}−\mathrm{32}{x}+\mathrm{64}=−\mathrm{14}{x}+\mathrm{49}−\mathrm{48}{x}+\mathrm{144} \\ $$$$−\mathrm{30}{x}+\mathrm{65}=−\mathrm{62}{x}+\mathrm{193} \\ $$$$\mathrm{32}{x}=\mathrm{128}\Rightarrow{x}=\mathrm{4} \\ $$$${Therefore}\:{center}\:{of}\:{the}\:{required}\:{circle}\:{is} \\ $$$$\left({x},\mathrm{11}−\mathrm{2}{x}\right)=\left(\mathrm{4},\mathrm{11}−\mathrm{2}×\mathrm{4}\right)=\left(\mathrm{4},\mathrm{3}\right) \\ $$$${Radius}\:{will}\:{be}\:{distance}\:{between}\:\left(−\mathrm{1},\mathrm{3}\:\right)\:{and}\:\left(\mathrm{4},\mathrm{3}\right) \\ $$$$\sqrt{\left(\mathrm{4}+\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{3}−\mathrm{3}\right)^{\mathrm{2}} }=\mathrm{5} \\ $$$$ \\ $$$${The}\:{equation}\:{of}\:{required}\:{circle} \\ $$$$\:\:\:\:\left({x}−\mathrm{4}\right)^{\mathrm{2}} +\left({y}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{5}^{\mathrm{2}} =\mathrm{25} \\ $$$${Or}\:{another}\:{form} \\ $$$$\:\:\:\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{8}{x}−\mathrm{6}{y}=\mathrm{0} \\ $$$$ \\ $$

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