Does Mathematica have a command that can directly get the center coordinates $(x,y)$ and radius $r$ of a circle?

Question

I have an equation of a circle:

x^2+y^2+2x-15==0

Does Mathematica have a command that can directly get the center coordinates $(x,y)$ and radius $r$?

Answer 1

c = ImplicitRegion[x^2 + y^2 + 2 x - 15 < 0, {x, y}];
center = RegionCentroid[c]
radius = Sqrt[Area[c]/Pi]

{-1, 0}

4

Note this does not verify if the input is actually a circle.

Answer 2

No, but you can easily implement your own function:

f[formula_] := {{a, b}, r} /. 
  Solve[{-2 a == Coefficient[formula, x], -2 b == Coefficient[formula, y], 
         a^2 + b^2 - r^2 == Last@MonomialList[formula], r > 0}, {a, b, r}][[1]]

so that

eq = x^2 + y^2 + 2 x - 15;

f[eq]

{{-1, 0}, 4}

Answer 3

Method 1:

circleForm = (x - a)^2 + (y - b)^2 - r^2;
sol = SolveAlways[x^2 + y^2 + 2 x - 15 == circleForm, {x, y}];
Pick[sol, NonNegative[r /. sol]]
(*  {{a -> -1, b -> 0, r -> 4}}  *)

As a function:

centerRadius[equation_, {x_, y_}] := Module[{res, a, b, r},
   res = SolveAlways[
     (equation /. Equal -> Subtract) == (x - a)^2 + (y - b)^2 - r^2,
     {x, y}];
   (* add check/message if SolveAlways fails, if desired *)
   ({{a, b}, r} /. Pick[res, NonNegative[r /. res]]) /; res =!= {}
   ];

centerRadius[x^2 + y^2 + 2 x - 15 == 0, {x, y}]
(*  {{{-1, 0}, 4}}  *)

Method 2: The gradient vanishes at the center; value of form at center is ±1 times the square of the radius, depending on the sign of the coefficients of the quadratic terms.

eqn = x^2 + y^2 + 2 x - 15 == 0;
center = First@Solve[D[eqn, {{x, y}}]]
radius = Sqrt[-(eqn /. Equal -> Subtract /. center)]  (* mult. by -1 * sign(coeff(x^2)) *)
(*
  {x -> -1, y -> 0}
  4
*)

Circle check (for checking function arguments, if desired):

circleQ[form_, {x_, y_}] := With[{ca = CoefficientArrays[form, {x, y}]},
    TrueQ[Length[ca] == 3] && 
     MatchQ[Normal@ca[[3]], {{a_, 0}, {0, a_}}]
    ];

circleQ[x^2 + y^2 + 2 x - 15, {x, y}]
(*  True  *)