Math

I need to write the standard equation for the circle that passes through the points:

(-5,0) (0,4) (2,4)

I have to use the general equation for a circle:

x^2+y^2+dx+ey+f=0

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  1. The center will lie at the intersection of the perpendicular bisectors of the line segments (chords) joining the points.

    If we label the points A,B,C, then

    The midpoint of AB is at (-5/2,2)
    The slope of the chord AB is 4/5
    The line AB is y = 4/5 (x+5)
    The slope of the radius to AB is thus -5/4
    The equation of the radius to AB is y = -5/4(x+5/2)+2

    The midpoint of BC is at (1,4)
    The slope of the chord BC is 0
    The line BC is y=4
    The radius to BC is vertical
    The equation of the radius to BC is x=1

    The midpoint of AC is at (-3/2,2)
    The slope of the chord AC is 4/7
    The line AC is y = 4/7(x+5)
    The slope of the radius to AC is thus -7/4
    The equation of the radius to AC is y = -7/4(x+3/2)+2
    The radii intersect at (1,-19/8), which is the center of the circle.

    The distance from the center to (0,4) is r^2 = 1+2601/64 = 2665/64

    So, the circle is

    (x-1)^2 + (y + 19/8)^2 = 2665/64

    You can see the circle and the chords at

    http://www.wolframalpha.com/input/?i=plot+%28x-1%29^2+%2B+%28y+%2B+19%2F8%29^2+%3D+2665%2F64%2C+y+%3D+4%2F5+%28x%2B5%29%2C++y%3D4%2C+y+%3D+4%2F7%28x%2B5%29&dataset=&equal=Submit

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    posted by Steve
  2. sketch a graph first

    If it goes through (0,4) and (2,4) I suspect that the center is on the line x = 1

    (x-1)^2 + (y-b)^2 = r^2
    when x = 0, y = 4
    1 + (4-b)^2 = r^2

    when x = 2, y = 4
    1 + (4-b)^2 = r^2 again of course

    when x = -5, y = 0
    36 + b^2 = r^2

    so
    1 +(4-b)^2 = b^2+36
    1 + 16 -8b + b^2 = b^2 + 36
    8 b = 17 - 36 = -19
    b = -19/8
    so
    (x-1)^2 + (y+19/8)^2 = 36 + 361/64 = 2665/64

    (x-1)^2 + (y+19/8)^2 = 6.45^2
    check my arithmetic and you can put that in your form if you choose

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    posted by Damon
  3. LOL !!!

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    posted by Damon
  4. Or, knowing that the general circle is
    (x-h)^2 + (y-k)^2 = r^2
    plug in your points to get
    (-5-h)^2 + k^2 = r^2
    h^2 + (4-k)^2 = r^2
    (2-h)^2 + (4-k)^2 = r^2

    h = 1
    k = -19/8
    r^2 = 2665/64
    and we have the circle as derived above

    The brute-force way, of course, is to plug your points into
    x^2+y^2+dx+ey+f=0

    (-5,0): 25-5d+f = 0
    (0,4): 16+4e+f = 0
    (2,4): 4+16+2d+4e+f = 0
    d,e,f = -2, 19/4, -35

    and the circle is thus

    x^2 + y^2 - 2x + 19/4 y - 35 = 0

    I'll let you verify that this is the same circle.

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    posted by Steve
  5. Let the circle pass through, in general, three points:
    (x1,y1)
    (x2,y2)
    (x3,y3)
    and the equation is to be expressed in the form

    x²+y²+dx+ey+f=0

    The required equation may be expressed in determinant form:
    [y^2+x^2, x, y, 1]
    [y1^2+x1^2, x1, y1, 1]
    [y2^2+x2^2, x2, y2, 1]
    [y3^2+x3^2, x3, y3, 1]

    When the determinant is expanded and equated to zero, the equation of the given circle in the required (polynomial) form will result.

    For the given three points (-5,0),(0,4),(2,4),
    the determinant expands to:
    x²+y²-2x+(19/4)y-35=0

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    posted by MathMate
  6. This is more fun every hour! Four ways now :)

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    posted by Damon

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