# Equation of a circle problem

• DeanBH
In summary, x^2 + y^2 + 2x -12y + 12 = 0 has an incorrect constant term, and therefore cannot be solved.

#### DeanBH

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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Your constant term is wrong. The center is right, but I got 5 as the radius. Try again and show your work.

accidental post

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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 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.

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

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.

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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nvm i understand it now

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,$r^2=f^2+g^2-c$ is $x^2+y^2=2fx+2gy+c=0$.
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.