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Equation of a circle problem

  • Thread starter DeanBH
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  • #1
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x^2 + y^2 + 2x -12y + 12 = 0

(x-a)^2 + (y-b)^2 = r^2 in that form.

why doesnt this work.


(x+1) + (y-6)

this gives you your x^2 your 2x and a +1 Gives you y^2 and -12y and a +36

so (x+1) + (y-6) + 49 = 0?

that would make r^2 negative which is impossible, what am i screwing up here?
 
Last edited:

Answers and Replies

  • #2
Vid
401
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Your constant term is wrong. The center is right, but I got 5 as the radius. Try again and show your work.
 
  • #3
82
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accidental post
 
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  • #4
82
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i have shown my working, i just tried to factorize in a way that would give me the co-efficients of x and y properly,

x^2 + y^2 + 2x -12y + 12 = 0

(x-a)^2 + (y-b)^2 = r^2 in that form.

why doesnt this work.


(x+1) + (y-6)

this gives you your x^2 your 2x and a +1 Gives you y^2 and -12y and a +36

i did (x + 1) because it gives you the 2x and the x^2 of the equation, but it also gives +1, i kept this on the left side of the equation so i ended up with +13 there. i then did the same for (y-6) and ended up with 36+13 on the left side of the equation which is 49.
 
  • #5
82
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when i multiply out the numbers in the brackets, do i have to put them over the right side of this equation, without changing the sign? as it were
 
  • #6
HallsofIvy
Science Advisor
Homework Helper
41,795
925
i have shown my working, i just tried to factorize in a way that would give me the co-efficients of x and y properly,

x^2 + y^2 + 2x -12y + 12 = 0
which is the same as x^2+ 2x+ y^2- 12y= -12

(x-a)^2 + (y-b)^2 = r^2 in that form.

why doesnt this work.


(x+1) + (y-6)
?? Did you mean (x+1)^2+ (y- 6)^2?

this gives you your x^2 your 2x and a +1 Gives you y^2 and -12y and a +36

i did (x + 1) because it gives you the 2x and the x^2 of the equation, but it also gives +1, i kept this on the left side of the equation so i ended up with +13 there. i then did the same for (y-6) and ended up with 36+13 on the left side of the equation which is 49.
Did you notice that you never do say what equation you wound up with?

Okay, (x+ 1)^2+ (y- 6)^2= x^2+ 2x+ 1+ y^2- 12y+ 36= what?
Since, in order to "complete the square" with x^2+ 2x, you have to add 1, and, in order to "complete the square" with y^2- 12y, you have to add 36, you have to add add a total of 37.
You started with x^2+ 2x+ y^2- 12y= -12. Adding 37 to both sides of that gives
(x+ 1)^2+ (y- 6)^2= 37-12= 25. You have -12+ 1+ 36, not 12+ 36+ 1, because they are on the right side of the equation.

The other way you have done this was, with your original equation, x^2+ y^2+ 2x- 12y+ 12= 0 is to recognize that you need a total of 36+ 1= 37 on left side. Since you already have 12, you need to add 37- 12= 25 to both sides:
x^2+ y^2+ 2x- 12y+ 12+ 25= 25 or (x^2+ 2x+ 1)+ (y^2- 12y+ 36)= (x+1)^2+ (y- 6)^2= 25.
 
  • #7
82
0
cant i do this without completing the square.

im just going to factorize, and whatever numbers i get move to the right side of the equation. lol.

that way it works even if i dont understand it
 
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  • #8
82
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nvm i understand it now
 
  • #9
rock.freak667
Homework Helper
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cant i do this without completing the square.

im just going to factorize, and whatever numbers i get move to the right side of the equation. lol.

that way it works even if i dont understand it
OR you could recall that the equation of a circle with centre (-f,-g) and radius,r,[itex]r^2=f^2+g^2-c[/itex] is [itex]x^2+y^2=2fx+2gy+c=0[/itex].
Then put your equation in that form and you'll get the centre and radius.
 

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