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Find the mid point of the chord 3x-y-1=0 to the circle x 2 +y 2 -4x+6y+1=0

Find the mid point of the chord 3x-y-1=0 to the circle x2+y2-4x+6y+1=0

Grade:11

1 Answers

kkbisht
90 Points
5 years ago
The circle x2 + y2 -4x+6y+1=0 and the chord 3x-y-1=0 intersect  at points given by x2 +(3x-1)2 -4x +6(3x-1) +1=0
=> x2 +9x2 +1 -6x -4x + 18x -6 +1=0 =>10x2 +8x-4=0 => 5x2 + 4x-2=0 => x= {-4 +-\sqrt{}16-4.5.(-2)}/10
x={-2+-2\sqrt{}14}/5
There fore y=3(-2+-2\sqrt{}14)/5) -1= ( -6+-6\sqrt{}14  +-5)/5 = (-11 +-6\sqrt{}14)/5
hence the mid point is x= ( x1+x2)/2={(-11 +2\sqrt{}14)/5 + (-11-2\sqrt{}14)/5}/2=-2/5
y= (y1 + y2)/2  =(-22/5)/2=-11/5
So mid point is : x=-2/5 , y=-11/5
 
K.K.Bisht
 

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