SOLUTION: one question: about finding center and radius (circle graph) : {{{ 3x^2 + 3y^2 - 6x + 12y= 0 }}} i did this: (3x^2 -6x) + (3y^2 +12y)=0 then I complete the square

Algebra ->  Graphs -> SOLUTION: one question: about finding center and radius (circle graph) : {{{ 3x^2 + 3y^2 - 6x + 12y= 0 }}} i did this: (3x^2 -6x) + (3y^2 +12y)=0 then I complete the square      Log On


   

Question 98840This question is from textbook
: one question: about finding center and radius (circle graph) :
+3x%5E2+%2B+3y%5E2+-+6x+%2B+12y=+0+


i did this:
(3x^2 -6x) + (3y^2 +12y)=0
then I complete the square:
-6/2 = -3^2=9 and 12/2 =6^2=36
then:
(3x^2-6x+9) + (3y^2+12y+36)= 9+36
(3x-3)^2 + (3y +6)^2 =45
factor:
3(x-1) + 3(y+2)= 45
i got the answer for center:
C(1,-2)
and the answer to the raduis is square root of 5
which i dont know how they got it.

thank you



This question is from textbook

Found 3 solutions by stanbon, scott8148, Earlsdon:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Find center and radius:
3x^2 + 3y^2 - 6x + 12y= 0
Divide thru by 3 to get:
x^2+y^2-2x+4y=0
Complete the square on the x-terms and y-terms separately:
x^2-2x+1 + y^2+4y+4 = 1+4
(x+1)^2 + (y+2)^2 = 5
--------------
Center at (-1,-2)
Radius = sqrt(5)
===============
Cheers,
Stan H.

Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
the first step in completing the square should be to make the coefficient of the squared term 1

in this case, the first step is to factor (reduce) if possible ... dividing by 3 gives x^2+y^2-2x+4y=0

completing the square gives x^2-2x+1+y^2+4y+4=5 ... (x-1)^2+(y+2)^2=5

if you FOIL (3x-3)^2, you don't get 3x^2-6x+9 ... similarly with (3y+6)^2

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Well, your error is found in the step..."completing the square"
Let's go over that:
3x%5E2-6x%2B3y%5E2%2B12y+=+0 then you said ..."complete the square", but before you do that, you should factor out a 3 then divide through by 3 and you will get:
3%28x%5E2-2x%2By%5E2%2B4%29+=+0 Divide both sides by 3.
x%5E2-2x%2By%5E2%2B4y+=+0 Now you complete the square when the coefficients of the x%5E2 and y%5E2 terms are equal to 1.
%28x%5E2-2x%2B1%29%2B%28y%5E2%2B4y%2B4%29+=+1%2B4 Factor the trinomials.
%28x-1%29%5E2%2B%28y%2B2%29%5E2+=+5 Now compare this with the standard form for a circle of radius r and center at (h, k).
%28x-h%29%5E2%2B%28y-k%29%5E2+=+r%5E2and you can see that h = 1, k = -2, and r%5E2+=+5
The center (h, k) is at (1, -2) and the radius is r+=+sqrt%285%29