Equation of a circle problem

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In summary, x^2 + y^2 + 2x -12y + 12 = 0 has an incorrect constant term, and therefore cannot be solved.
  • #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 doesn't 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?
 
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  • #2
Your constant term is wrong. The center is right, but I got 5 as the radius. Try again and show your work.
 
  • #3
accidental post
 
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  • #4
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 doesn't 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
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
DeanBH said:
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 doesn't 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
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 don't understand it
 
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  • #8
nvm i understand it now
 
  • #9
DeanBH said:
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 don't 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.
 

1. What is the equation of a circle?

The equation of a circle is given by (x - h)^2 + (y - k)^2 = r^2, where (h,k) represents the center of the circle and r represents the radius.

2. How do you find the center of a circle using the equation?

The center of a circle can be found by setting the values of x and y to 0 in the equation (x - h)^2 + (y - k)^2 = r^2. This will give the coordinates (h,k) of the center.

3. Can you find the radius of a circle using the equation?

Yes, the radius of a circle can be found by taking the square root of the number on the right side of the equation, r^2. This will give the value of r, which is the radius.

4. How do you graph a circle using the equation?

To graph a circle using the equation, first find the center coordinates and the radius. Plot the center point on the coordinate plane and use the radius to draw the circle around it.

5. Is there a different equation for a circle if the center is not at the origin?

Yes, if the center of the circle is not at the origin, the equation would be (x - a)^2 + (y - b)^2 = r^2, where (a,b) represents the coordinates of the center. The process for finding the center and radius remains the same.

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