Question 816642
Here we are looking for a circle with its center at the midpoint between
 (0,-1) and (1/3,4/3).  The midpoint is ( (0+1/3)/2 , (-1+4/3)/2 ).
  ( 1/6 , 1/6 ).  We also need the length of the radius of the circle
  which has as diameter (0,-1) and (1/3,4/3).  The length of the diameter
  is sqrt( (y2-y1)^2 + (x2-x1)^2 ).
  So we want sqrt ( ( 1/3 - 0 )^2 + ( 4/3 - (-1) )^2 )
  = sqrt ( ( 1/3 )^2 + ( 4/3 + 1 )^2 )
  = sqrt (  1/9 + (7/3)^2 )
  = sqrt (  1/9 + 49/9 )
  = sqrt ( 50/9 )
  The radius is half of this or 25/9 .

  The standard form for a circle is (x-h)^2 + (y-k)^2 = r^2
  For our application h= 1/6 and k= 1/6 , with r = 25/9 r^2 = 625/81
  So (x- 1/6)^2 + (y- 1/6)^2 = 625/81

  Double check this.  Plotting the above equation
  http://www.wolframalpha.com/input/?i=%28x-+1%2F6%29^2+%2B+%28y-+1%2F6%29^2+%3D+625%2F81