Another take. This is a function that takes a polynomial in any number of variables. Where all variables appear to the 2nd power, and not higher. The 2nd power terms has to have coefficients of +1:

completeTheSquare[poly_] := With[
  {form = Total@Table[(i + Unique[])^2, {i, Variables@poly}] + Unique[]},
  form /. Flatten@Solve[ForAll[Evaluate@Variables@poly, poly == form]]
] /;
And[
  Union[Exponent[poly, Variables@poly]] == {2},
  Union[Coefficient[poly, #, 2] & /@ Variables@poly] == {1}
]

Applying the function:

completeTheSquare[t^2 + 10 t - 39]
(* -64+(5+t)^2 *)

completeTheSquare[x^2 + y^2 - 2 x + 6 y + 9]
(* -1+(-1+x)^2+(3+y)^2 *)

completeTheSquare[i^2 + j^2 + k^2 - 2 i - 4 j - 6 k - 11]
(* -25+(-1+i)^2+(-2+j)^2+(-3+k)^2 *)

You could as well do it directly using elementary Mma-commands

gl = x^2 + y^2 - 2 x + 6 y + 9;

pp = {#[[2, 1]], #[[1, 2]], #[[1, 1]]} &[CoefficientList[gl, {x, y}]];

res = Factor[gl - pp[[3]] + pp[[1]]^2/4 /. y -> 0] +
  Factor[gl - pp[[3]] + pp[[2]]^2/4 /. x -> 0] -
  pp[[1]]^2/4 - pp[[2]]^2/4 + pp[[3]]

You did not load Parks's Presentations package, which you must first obtain, install, and then load:

<<Presentations`

Hi

There is a catch: The method requires coefficients of the $x^2$ and $y^2$ terms to be +1.

So in your case divide the expression by 4. This will give the expected output.

template = (x-a)^2 + (y-b)^2 + c

expression = 4 x^2 + 4 y^2 + 12 x - 24 y + 41

template /. First@Solve[ForAll[{x, y}, expression/4 == template]]
(* -1 + (3/2+x)^2 + (-3+y)^2 *)

Mitchell,

As you have probably already noted this method can also be used in general to "complete the square"

template = (x - a)^2 + b

expression = x^2 + 10 x - 39

z = Solve[ForAll[x, expression == template]]

template /. First@z // TraditionalForm
(x + 5)^2 - 64

Hi Hans;

That is a neat example. However when I tried it on my system, I did not get the same results - see attached.

Attachments:

HansExample.nb

Probably you what:

c = ImplicitRegion[x^2 + y^2 - 2 x + 6 y + 9 < 0, {x, y}];
center = RegionCentroid[c]
radius = Sqrt[Area[c]/Pi]

(*{1, -3}*)
(* 1 *)

For more solution for this problem see here.

Regards M.I.