# Simultaneous solving equations

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?

• 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]. – Victoria Oct 17 at 23:54
• What have you tried? – David G. Stork Oct 18 at 0:45
• I don't understand how to set it up. – Victoria Oct 18 at 0:50
• This looks like a homework question. A hint: there are 3 unknowns a, b, R, and three points, solve the equations. – Rohit Namjoshi Oct 18 at 18:21

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

SeedRandom

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,
Tooltip[Point[#], #] & /@ pts}],
AxesOrigin -> ({a, b} /. sol),
Axes -> True,
PlotLabel -> sol]
` 