College Algebra

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using the knowledge of systems of equations, and that the equation of a circle is (x-h)^2 + (y-k)^2 = r^2, where (h,k) is the center and r is the radius. Find the equation of a circle that contains the following points: (6,2), (4,6), and (-3,5). (These 3 points each create an equation. Then you have a system. Use ELIMINATION to solve.)

  • College Algebra -

    The method suggested for your question is not the best one in this case.
    I will use the property that the centre lies on the intersection of the right - bisector of any two chords.

    for (6,2) and (4,6)
    slope = (6-2)/4-6) = -2
    so slope of perpendicular is +1/2
    midpoint of chord = ( (6+4)/2 , (2+6)/2 ) = (5,4)
    equation of rt-bisector : y =(1/2)x + b
    but (5,4) lies on it
    4 = (1/2)(5) + b
    b = 4-5/2 = 3/2
    b = 3/2 --------> y = (1/2)x +3/2

    for (6,2) and (-3,5)
    slope of line = (5-2)/(-3-6) = -1/3
    slope of perp is 3
    midpoint of chord = (3/2, 7/2)
    y = 3x + b
    but (3/2,7/2) lies on it
    7/2 = 3(3/2) + b
    b = 7/2 - 9/2 = -1 ---- y = 3x - 1

    intersection: 3x-1 = (1/2)x+3/2
    times 2
    6x - 2 = x + 3
    5x = 5
    x = 1
    then y = 3(1) - 1 = 2
    so we know the centre is (1,2)
    using (6,2),
    r^2 = (6-1)^2 + (2-2)^2
    = 25

    circle equation: (x-1)^2 + (y-2)^2 = 25

    check:
    for (6,2) --- LS = 25 + 0 = 25✔
    for (4,6) --- LS = 9 + 16 = 25✔
    for (-3,5) -- LS = 16 + 9 = 25✔

  • College Algebra -

    However, just to see how elimination works, plug in the points and solve the 3 equations. The trick is how to get rid of those pesky squared terms. Just equate the various values of r^2.

    (6-h)^2 + (2-k)^2 = r^2
    (4-h)^2 + (6-k)^2 = r^2
    (-3-h)^2 + (5-k)^2 = r^2

    (6-h)^2 + (2-k)^2 = (4-h)^2 + (6-k)^2
    (6-h)^2 + (2-k)^2 = (-3-h)^2 + (5-k)^2
    (4-h)^2 + (6-k)^2 = (-3-h)^2 + (5-k)^2

    36-12h+h^2 + 4-4k+k^2 = 16-8h+h^2 + 36-12k+k^2
    36-12h+h^2 + 4-4k+k^2 = 9+6h+h^2 + 25-10k+k^2
    16-8h+h^2 + 36-12k+k^2 = 9+6h+h^2 + 25-10k+k^2

    36-12h + 4-4k = 16-8h + 36-12k
    36-12h + 4-4k = 9+6h + 25-10k
    16-8h + 36-12k = 9+6h + 25-10k

    -4h +8k = 12
    -18h + 6k = -6
    -14h - 2k = -18

    Since we only have h and k, we only need two equations:

    -h + 2k = 3
    -6h + 2k = -2
    5h = 5
    h = 1
    k = 2

    (x-1)^2 + (y-2)^2 = r^2
    pick any point:
    (6-1)^2 + (2-2)^2 = 25 = r^2
    r = 5

    (x-1)^2 + (y-2)^2 = 25

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