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Question about circle

  1. Mar 23, 2010 #1
    hi

    1. The problem statement, all variables and given/known data

    determine if whether it's circle or not , if yes find the center and radius

    1) 2x^2 + 2y^2 - 5x + 4y - 1 =0

    2) x^2 + y^2 + 4x + 8y + 25 =0



    3. The attempt at a solution

    1) center (5/4,-1)
    2) center (2,4)

    well i need help in how to find the " radius " and how to know if it's circle equation or not ,,,
     
  2. jcsd
  3. Mar 23, 2010 #2

    rock.freak667

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    If it is a circle, it can be put into the form

    (x-a)2+(y-b)2=r2

    with center (a,b) and radius 'r'.
     
  4. Mar 23, 2010 #3
    i know that ,,,, but now it's expanded ,,,

    so i will got b^2 + a^2 - r^2 in one side ,,,,,

    how i will know if it's circle and the " r " ?
     
  5. Mar 23, 2010 #4

    rock.freak667

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    Gather all the terms with 'x' and compete the square, do the same for the 'y', then move all the constants to the right side of the equation and that will be your r2
     
  6. Mar 23, 2010 #5
    i completed the square and i got the points ,,, but how to know if it's circle equation ? because #2 is not ,,,
    but i dont know why ,,,,
     
  7. Mar 23, 2010 #6
    The general equation of circle are in the form of x^2+y^2+2gx+2fy+c=0 and since the equations that you mentioned are in this form I am pretty sure they're a circle.
     
  8. Mar 23, 2010 #7
    2) x^2 + y^2 + 4x + 8y + 25 =0
    is not circle equation am sure because I see it in my book ,,,,
    and I don't know why it's not ,,,
     
  9. Mar 23, 2010 #8
    No it's not a circle ,, Put "x^2 + y^2 + 4x + 8y + 25 =0 " in circle equation :
    (x+2)^2+(y+4)^2=-20 ,, the radius is minus so it's not a circle :smile:
     
  10. Mar 23, 2010 #9
    aha ,,,, thx Lord
     
  11. Mar 23, 2010 #10
    no problem :biggrin:
     
  12. Mar 24, 2010 #11
    Would you please elaborate why?
     
  13. Mar 24, 2010 #12
    Why what ? it's not a circle ?? ,, if yes here's why :
    the equation
    x^2 + y^2 + 4x + 8y + 25 =0
    >> (x^2+4x+...)+(y^2+8y+...) = -25 [complete the square for both of them]
    >> (x+2)^2+(y+4)^2 = -25+4+16 [add on both sides 4 & 16 to be able to complete the square]
    >>(x+2)^2+(y+4)^2= -5 [ you get the radius -5 which tells you that there is no circle with minus radius]

    hope you understand now ,, and sorry for the wrong answer before (r=-20)
     
  14. Mar 24, 2010 #13
    I don't understand why a circle can't have negative radius. For example, consider this circle with equation x^2+y^2=0, which has its center at origin and passes through points P(x, 0) & Q(-x, 0). If the radius of a circle is defined as the distance from the center to any point on the circle, then the circle in the above example would have negative radius in the 2nd and 3rd quadrant. Thanks for the reply in advance =)
     
  15. Mar 24, 2010 #14

    rock.freak667

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    Distances aren't negative.
     
  16. Mar 24, 2010 #15

    LCKurtz

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    [itex](x-a)^2 + (y-b)^2 \ge 0[/itex]. You can't have it equal to a negative number and find any (x,y) that work.
     
  17. Mar 24, 2010 #16

    Mark44

    Staff: Mentor

    The equation x^2 + y^2 = 0 represents a degenerate circle, a circle whose radius is 0. This "circle" DOES NOT pass through points P(x, 0) and Q(0, y), unless x = 0 and y = 0.
     
  18. Mar 24, 2010 #17
    @rock.freak667: I know but I got a little confused by the structure of the equation.
    @LCKurtz: Thanks that helps me understand the concept better :)
    @Mark44: I actually though about that after posting :D

    Thanks for the help guys =)
     
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