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    Find where the circle (x-5)^2 + (y+1)^2 = 9 meets the x-axis. My brain is mal-functioning...

    Edit: I was trying to solve for X=0 instead of Y=0. Thanks Femto.
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    (Original post by Chemist548)
    Find where the circle (x-5)^2 + (y+1)^2 = 9 meets the x-axis. My brain is mal-functioning...
    lol, feels like aaaages since i did maths but if i remember correctly:

    circle meats x axis when x=0
    x=0 when the x bracket=0
    x bracket=0 when x = +5
    so x = 5

    if you have to find the y coordinate, put x=5 into the equation and juggle the numbers around (with all like terms (numbers without y term) on the same side). it should come to y^2 = ...., then just square root both sides to give y = ....

    i dont wana work it out for you bcoz i want you to understand how to do it
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    When it meets the x axis, y = 0

    Can you work it out from there?
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    (Original post by Femto)
    When it meets the x axis, y = 0 (Ash92 you may have read the question wrong).

    Can you work it out from there?
    oooo, of course. thanks for the correction. i'll try to change the post and resend it
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    (Original post by Femto)
    When it meets the x axis, y = 0

    Can you work it out from there?
    Oh for christ sake, I had X=0 instead of Y=0 and that's why I couldn't solve the quadratic.

    Thanks
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    • circle meats x axis when y=0
    • when y=0, the y bracket=0 [so (y+1)^2 = (0+1)^2 = 1]
    • so "(x-5)^2 + (y+1)^2 = 9" = "(x-5)^2 + 1 = 9"
    • now juggle the numbers as i said before to have any x terms on the same side (x^2) and the numbers on the other side of the equation.
      so (x^2 = .....)
    • now square root both sides


    sorry for the confusion
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    (Original post by Chemist548)
    Oh for christ sake, I had X=0 instead of Y=0 and that's why I couldn't solve the quadratic.

    Thanks
    lol, i made same mistake.
 
 
 
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