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Analytical Geometry> Find the mid point of the chord 3x-y-1=0 ...

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

HIMADRI BISWAS , 7 Years ago

Grade 11

1 Answers

kkbisht

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+x_2⁾/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

Last Activity: 7 Years ago

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