Algebra

posted by Lacey

X^2+y^2+6y+3=0
x^2+y^2-9x-6=0

List the solutions:

Please and thank you!

  1. Steve

    You have two circles:

    x^2 + (y+3)^2 = 6
    (x - 9/2)^2 + y^2 = 105/4

    Naturally, they intersect where

    x^2+y^2+6y+3 = x^2+y^2-9x-6
    6y = -9x-9
    y = -(3x+3)/2

    Subbing that into one of the equations,
    x^2+(3x+3)^2/4-9x-6=0
    13x^2 - 18x - 15 = 0
    x = 1/13 (9±2√69)
    y = -3/13 (11±√69)

    Hmm. I had expected a bit simpler answer.

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