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Find thev equation of chord of circle x^2+y^2=4x which is bisected at the point (1,1)

Find thev equation of chord of circle x^2+y^2=4x which is bisected at the point (1,1)

Grade:

2 Answers

Vikas TU
14149 Points
7 years ago

equation of chord =

xx1 + yy1 -4(x + x1)/2 = 0

Pass the eqn. from (1,1)

x + y - 2x - 2 = 0

or

y = x + 2 

 

Approve plz!

vikram bedi
4 Points
7 years ago

Thnxxx but How did u change 4x to 4(x+x1)/2????

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