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Analytic Geometry I - Circle

  1. Nov 20, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the centre and radius of the circle x²+y²-4x+2y+6=0

    I have the solution. The circle is no defined because r² = -1 is impossible.
    But... how do I even DO that equation to get the answer -1?!
     
  2. jcsd
  3. Nov 20, 2008 #2

    Dick

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    You complete the squares so you can write it in the form (x+a)^2+(y+b)^2=c. Do you know how to do that?
     
  4. Nov 20, 2008 #3
    I learned how to do that while I was in school.. I forget how to do it now.. >_<
     
  5. Nov 20, 2008 #4

    Dick

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    Take the x part. You've got x^2-4x. If I add something to that it will become a perfect square of the form (x-a)^2. What's 'a'? What do you have to add?
     
  6. Nov 20, 2008 #5
    Oh! Is "a" 2x?
    (x-2x)² ?
     
  7. Nov 20, 2008 #6

    Dick

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    No, no. (x-a)^2=x^2-2ax+a^2, yes? If you match that up with x^2-4x, the 4 must be the 2a, as I see it. Think back to when you did this before.
     
  8. Nov 20, 2008 #7
    Ohh, I see now.. so I just do the same thing for y?
     
  9. Nov 20, 2008 #8

    Dick

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    Sure.
     
  10. Nov 20, 2008 #9
    Thank you so much! :)
     
  11. Nov 21, 2008 #10
    it can actually be express as x²+y²+2fx+2gy+c=0

    whrby C(-f,-g) and radius is [tex]\sqrt{g²+f²-c}[/tex]
     
  12. Nov 21, 2008 #11

    HallsofIvy

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    No, it isn't. It is much better to actually do the "complete the square" rather than memorize formulas: so you don't make silly mistakes like that.
     
  13. Nov 21, 2008 #12
    yep (: noted.
     
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