Solid mensuration

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Find an equation of the circle tangent to x + y = 3 at (2,1) and with center on 3x - 2y - 6 = 0

  • Solid mensuration -

    y = - x + 3 so
    slope of tangent = -1
    slope of line from (2,1) to circle center = -1/-1 = 1 because radius of circle is perpendicular to tangent at intersection
    so find the radius line to the tangent point (2,1)
    y = +1 x + b
    1 = 2 + b
    b = -1
    so that radius line is y = x-1
    the center is also on 3 x - 2 y = 6
    so
    3 x - 2 (x-1) = 6
    x = 4
    y = 3
    center at (4,3)
    so form is
    (x-4)^2 + (y-3)^2 = r^2
    we still need r^2 but we know center at (4,3) and tangent point at (2,1)
    distance between is r
    r^2 = (1-3)^2 + (2-4)^2
    r^2 = 4 + 4 = 8
    so
    (x-4)^2 + (y-3)^2 = 8
    check my arithmetic, I did that fast.

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