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Math q.

  1. Nov 9, 2003 #1
    point A had the coordinates (8,1) and B(7,0)
    find the equation of the circle passing through A and B.

    hope to get rough idea on how to solve this question. tq.
     
  2. jcsd
  3. Nov 9, 2003 #2

    jamesrc

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    general equation for a circle:

    (x-xc)^2 + (y-yc)^2 = R^2

    where xc is the x-coordinate of the center of the circle, yc is the y-coordinate of the center of the circle, and R is the radius of the circle.

    To fully define a circle, you need to specify 3 points on its circumference (unless you, for example, know the location of the center). This is borne out in the equations: If you plug in the known info to this equation:

    point A:

    (8-xc)^2 + (1-yc)^2 = R^2

    point B:

    (7-xc)^2 + (0-yc^2) = R^2


    You have 2 equations and 3 unknowns, so you don't have a uniquely defined circle. If you know xc, yc, or R, you can proceed and write out the equation for the circle.
     
  4. Nov 9, 2003 #3
    here's the actual question.

    point A had the coordinates (8,1)
    and B (7,0). find the equation of the circle passing through A and B, and its tangent at the point B had the equation 3x-4y-21=0.
    find the equation of the tangent parallel with the tangent at B.

    can it be solved?
     
  5. Nov 9, 2003 #4

    jamesrc

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    Yes, it is solvable.

    Get your third equation from the slope of the line:

    first write the equation of the line in standard form:
    3x-4y-21=0 --> y = .75x-(21/4)

    Now find an expression for the slope from the equation for the circle:

    (x-xc)^2 + (y-yc)^2 = R^2

    use implicit differentiation to get:

    2(x-xc)dx + 2(y-yc)dy = 0 --> (dy/dx) = -(x-xc)/(y-yc)

    Evaluate this at point B so you can equate with the slope above:

    (dy/dx)[at B] = -(7-xc)/(0-yc) = 0.75

    so you have your 3 equations:

    xc = 7 - .75yc
    (8-xc)^2 + (1-yc)^2 = R^2
    (7-xc)^2 + (0-yc^2) = R^2

    and you can use those to find xc, yc, and R
     
  6. Nov 9, 2003 #5
    thank you.
     
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