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Solving equation of a circle in Mathematica

Mitchell Sandlin
Posted 6 years ago

How would you go about solving the function: x^2+y^2- 2x + 6y + 9 = 0 in Mathematica? This function has the solution of: (x-1)^2+(y+ 3)^2 = 1, which is a circle of radius 1 centered at (1,-3).

When using the Solve[] function of: 

Solve[x^2 + y^2 - 2 x + 6 y + 9 == 0 {x, y}]
 I get {{y -> -3 - Sqrt[2 x - x^2]}, {y -> -3 + Sqrt[2 x - x^2]}}

which is far from my desired answer.

I know I am omitting something from the function, but cannot figure out what.

Thanks,

Mitch Sandlin
POSTED BY: Mitchell Sandlin
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Use the function CompleteTheSquare from the Wolfram Function Repository: https://resources.wolframcloud.com/FunctionRepository/resources/CompleteTheSquare
POSTED BY: Murray Eisenberg
Hans Milton
Hans Milton
Posted 6 years ago
POSTED BY: Hans Milton
POSTED BY: Hans Dolhaine

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]]
POSTED BY: Hans Dolhaine

Hi Murray;

I tried duplicating your example, but did not get the same results. Please see attached notebook and give me your thoughts.

Thanks so much,

Mitch Sandlin

Attachments:
CompleteTheSquare.nb
POSTED BY: Mitchell Sandlin

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

<<Presentations`
POSTED BY: Murray Eisenberg

One of the many useful functions, CompleteTheSquare, from David Parks's Presentations application, makes this almost immediate:

    poly = Fold[CompleteTheSquare, x^2 + y^2 - 2 x + 6 y + 9, {x, y}]
(* -1 + (-1 + x)^2 + (3 + y)^2 *)

And then use the built-in AddSides:

    AddSides[poly == 0, -First@poly]
(* (-1 + x)^2 + (3 + y)^2 == 1 *)
POSTED BY: Murray Eisenberg
Hans Milton
Hans Milton
Posted 6 years ago

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 *)
POSTED BY: Hans Milton

Thanks Hans, and I did notice that the technique could be used for "Completing the Square", which I find very interested.

I did come up with one additional question concerning handling coefficient's to x^2 and y^2 - see attached. I am not quite sure how to handle these additional coefficients in the technique, which is causing Mathematica to generate computational errors. Completing the Square manually, I get: (x + 3/2)^2 + (y - 3)^2 =1

Thanks,

Mitch Sandlin

Attachments:
Hans2Example.nb
POSTED BY: Mitchell Sandlin
Hans Milton
Hans Milton
Posted 6 years ago

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
POSTED BY: Hans Milton

Hi Hans;

Please ignore my prior response because it was me at fault - I wasn't clearing the variables correctly. Your neat example works fine and thanks again.
POSTED BY: Mitchell Sandlin

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
POSTED BY: Mitchell Sandlin
Hans Milton
Hans Milton
Posted 6 years ago 
template = (x - a)^2 + (y - b)^2 + c

expression = x^2 + y^2 - 2 x + 6 y + 9

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

I picked this up from a MSE discussion. See the answer from minthao_2011, towards the end.
POSTED BY: Hans Milton

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.
POSTED BY: Mariusz Iwaniuk
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