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Question Number 170378 by Mastermind last updated on 22/May/22
The line 2y=x+3 meets the circle   x^2 +y^2 −2x−6y−15 at point M and N  where N is in the second quadrant.  find the coordinate of M and N.    Mastermind
$${The}\:{line}\:\mathrm{2}{y}={x}+\mathrm{3}\:{meets}\:{the}\:{circle}\: \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{6}{y}−\mathrm{15}\:{at}\:{point}\:{M}\:{and}\:{N} \\ $$$${where}\:{N}\:{is}\:{in}\:{the}\:{second}\:{quadrant}. \\ $$$${find}\:{the}\:{coordinate}\:{of}\:{M}\:{and}\:{N}. \\ $$$$ \\ $$$${Mastermind} \\ $$
Answered by daus last updated on 22/May/22
x^2 +y^2 −2x−6y=15  2y−3=x  (2y−3)^2 +y^2 −2(2y−3)−6y=15  4y^2 −12y+9+y^2 −4y+6−6y=15  5y^2 −22y=0  y(5y−22)=0  y=0 , 5y−22=0                    y=((22)/5) or 4.4  x=−3  , x=5.8   N(−3,0)  M(5.8,4.4)
$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{6}{y}=\mathrm{15} \\ $$$$\mathrm{2}{y}−\mathrm{3}={x} \\ $$$$\left(\mathrm{2}{y}−\mathrm{3}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{2}{y}−\mathrm{3}\right)−\mathrm{6}{y}=\mathrm{15} \\ $$$$\mathrm{4}{y}^{\mathrm{2}} −\mathrm{12}{y}+\mathrm{9}+{y}^{\mathrm{2}} −\mathrm{4}{y}+\mathrm{6}−\mathrm{6}{y}=\mathrm{15} \\ $$$$\mathrm{5}{y}^{\mathrm{2}} −\mathrm{22}{y}=\mathrm{0} \\ $$$${y}\left(\mathrm{5}{y}−\mathrm{22}\right)=\mathrm{0} \\ $$$${y}=\mathrm{0}\:,\:\mathrm{5}{y}−\mathrm{22}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{y}=\frac{\mathrm{22}}{\mathrm{5}}\:{or}\:\mathrm{4}.\mathrm{4} \\ $$$${x}=−\mathrm{3}\:\:,\:{x}=\mathrm{5}.\mathrm{8}\: \\ $$$${N}\left(−\mathrm{3},\mathrm{0}\right)\:\:{M}\left(\mathrm{5}.\mathrm{8},\mathrm{4}.\mathrm{4}\right)\: \\ $$

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