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)
vikram bedi , 12 Years ago
Grade
Analytical Geometry> pleazzz-answer...
2 AnswersVikas TU
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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! |
Approved Last Activity: 12 Years ago
vikram bedi
Thnxxx but How did u change 4x to 4(x+x1)/2????
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