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Analytic Geometry

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Find the equation of a circle passing through (3,7) and tangent to the line x-3y+8=0

  • Analytic Geometry - ,

    The the centre be C(a,b)
    and the point of contact with the line P(x,y)
    let A(3,7) be the given point on the circle

    AP must be a diameter, and at right angles to the given line
    slope of given line = 1/3
    so slope of diameter = -3
    but (3,7) lies on it, so the equation is
    y = -3x + b
    7 = -3(3) + b
    b = 16
    diameter has equation y = -3x + 16

    solving the two equations would give us P
    x - 3(-3x+16) + 8 = 0
    x + 9x - 48 + 8 = 0
    10x = 40
    x = 4
    then y = -3(4) + 16 = 4
    P is the point (4,4)
    so the centre must be the midpoint and
    C = (7/2, 11/2)

    circle equation:
    (x-7/2)^2 + (y-11/2)^2 = r^2
    r = 1/2 the diameter
    diameter = √( (-1)^2 + 3^2 ) = √10
    r = √10/2
    r^2 = 10/4 = 5/2

    circle equation:
    (x-7/2)^2 + (y-11/2)^2 = 5/2

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