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Find the equation of the smaller circle tangent to both axes and passing the point (-3,6)

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    If the circle is tangent to both axes and passes through (-3,6), I see its centre to be in the 2nd quadrant.
    If the radius is r, then its centre must be (-r,r) and its equation is
    (x+r)^2 + (y-r)^2 = r^2
    x^2 + 2rx + r^2 + y^2 - 2ry + r^2 = r^2
    at (-3,6)
    9 - 6r + r^2 + 36 - 12r = 0
    r^2 - 18r + 45 = 0
    (r-3)(r-15) = 0
    r = 3 or r = 15

    since you want the smaller circle ... r = 3
    equation:
    (x+3)^2 + (y-3)^2 = 9

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