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It is a theorem in geometry that any three non-collinear points determine a circle. Given the three points (x1,y1),(x2,y2),(x3,y3) find a,b,and R for the circle (x-a)^2 +(y - b)^2 =R that passes through them. Notice that the solutions have the same denominator or its square. What does the equation denominator=0 signify for the three points?

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    $\begingroup$ Hint: here, you want to solve 3 simultaneous equations for the variables {a, b, R}. For the first equation, you want the circle to pass through the point (x1, y1), so you replace x with x1 and y with y1 in the equation of the circle. To answer the last question, what is the situation between the 3 points for which it is impossible to obtain a circle? Note: use x0 rather than Subscript[x, 0]. $\endgroup$ – Victoria Oct 17 at 23:54
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    $\begingroup$ What have you tried? $\endgroup$ – David G. Stork Oct 18 at 0:45
  • $\begingroup$ I don't understand how to set it up. $\endgroup$ – Victoria Oct 18 at 0:50
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    $\begingroup$ This looks like a homework question. A hint: there are 3 unknowns a, b, R, and three points, solve the equations. $\endgroup$ – Rohit Namjoshi Oct 18 at 18:21
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Clear["Global`*"]

Format[x[n_]] := Subscript[x, n]
Format[y[n_]] := Subscript[y, n]

eqn = (x - a)^2 + (y - b)^2 == R^2;

eqns = eqn /. Transpose[Thread[# -> Array[#, 3]] & /@ {x, y}]

enter image description here

sol = Solve[Append[eqns, R > 0], {a, b, R}, Reals][[1]];

SeedRandom[1234]

pts = Evaluate@Array[{x[#], y[#]} &, 3] = RandomReal[{-10, 10}, {3, 2}]

(* {{7.53217, 0.439285}, {-8.27553, -2.44174}, {-9.76711, 8.54532}} *)

Show[
 ContourPlot @@ ({eqn, {x, a - R - 1, a + R + 1}, {y, b - R - 1, 
      b + R + 1}} /. sol),
 Graphics[{Red, AbsolutePointSize[6],
   Tooltip[Point[#], #] & /@ pts}],
 AxesOrigin -> ({a, b} /. sol),
 Axes -> True,
 PlotLabel -> sol]

enter image description here

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