Asked by Ande2
Find the centre of the circle 36x^2+36y^2-24x-36y-23=0
Answers
Answered by
oobleck
complete the squares and it becomes clear:
36x^2+36y^2-24x-36y-23=0
36x^2-24x + 36y^2-36y = 23
36(x^2 - 2/3 x) + 36(y^2-y) = 23
36(x^2 - 2/3 x + 1/9) = 36(y^2 - y + 1/4) = 23 + 36(1/9) + 36(1/4)
36(x - 1/3)^2 + 36(y - 1/2)^2 = 36
(x - 1/3)^2 + (y - 1/2)^2 = 1
36x^2+36y^2-24x-36y-23=0
36x^2-24x + 36y^2-36y = 23
36(x^2 - 2/3 x) + 36(y^2-y) = 23
36(x^2 - 2/3 x + 1/9) = 36(y^2 - y + 1/4) = 23 + 36(1/9) + 36(1/4)
36(x - 1/3)^2 + 36(y - 1/2)^2 = 36
(x - 1/3)^2 + (y - 1/2)^2 = 1
Answered by
olarinoye olamide
thanks alot
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