need help with another equation of a circle

[abel muroi]

I just did a problem that gave me a equation and asked me to find the center and radius of the equation.

here is the equation...

3x² + 3y² + 6x - y = 0

and here is the work i did...

3x² + 3y² + 6x - y = 0
x² + y² + 2x - y = 0
(x² + 2x) + (y² - y) = 0
(x² + 1) + (y² - y) = 1
(x - 1)² + (y² - 0)² = 1

center = (1,0)
radius = 1

I feel that i made a mistake somewhere since i wasn't too sure what i was supposed to do with the -y in the equation. can someone tell me what was my mistake here?

 

[Steven G]

I just did a problem that gave me a equation and asked me to find the center and radius of the equation.

here is the equation...

3x² + 3y² + 6x - y = 0

and here is the work i did...

3x² + 3y² + 6x - y = 0
x² + y² + 2x - y = 0 you did not divide the y by 3
(x² + 2x) + (y² - y) = 0
(x² + 1) + (y² - y) = 1 what happened to the 2x from the line above??
(x - 1)² + (y² - 0)² = 1 (x - 1)² is not (x² + 1). (y² - 0)² = (y² )² =y⁴ is not (y² - y)

center = (1,0)
radius = 1

I feel that i made a mistake somewhere since i wasn't too sure what i was supposed to do with the -y in the equation. can someone tell me what was my mistake here? -y = -1y. So you compute 1/2 of -1 and then you square it . This number is added to both sides of the equation

You need to follow the steps. After getting a 1 in front of the x^2 and y^2 terms you take half of the coefficient of the x and y term and then square it and....

 

[Ishuda]

I just did a problem that gave me a equation and asked me to find the center and radius of the equation.

here is the equation...

3x² + 3y² + 6x - y = 0

and here is the work i did...

3x² + 3y² + 6x - y = 0
x² + y² + 2x - y / 3 = 0 <======= you also need to divide -y by 3
(x² + 2x) + (y² - y) = 0
(x² + 1) + (y² - y) = 1
(x - 1)² + (y² - 0)² = 1

center = (1,0)
radius = 1

I feel that i made a mistake somewhere since i wasn't too sure what i was supposed to do with the -y in the equation. can someone tell me what was my mistake here?

You need to learn how to complete the square. If you remember from the other thread with the same kind of problem,
http://www.freemathhelp.com/forum/threads/90014-finding-radius-and-center-of-an-equation
I told you that the equation of a circle was
$\displaystyle (x-x_0)^2 + (y-y_0)^2 = r^2$
where (x₀, y₀) was the center of the circle and r was the radius. Jomo walked you through a couple of examples. I'll present it a little different way and maybe that will help.

First expand the equation for the circle and write it as
(a) $\displaystyle x^2 - 2 x_0 x + x_0^2 + y^2 - 2 y_0 y + y_0^2 = r^2$
Let's compare a quadratic equation with this. Rather than use yours which would be, rewriting it slightly,
x² + 2 x + y² - (1/3) y = 0
we will use
2 x² - 16 x + 2 y² - 4 y + 2 = 0
or, dividing it through by 2 as you did yours to get the standalone x² and y²,
(b) x² - 8 x + y² - 2 y + 1 = 0

Compare the x's for both equations (a) and (b):
(1) The coefficients of x² are the same; both are equal to one
(2) The coefficients of x are the same so -8 = -2 x₀ or x₀ = 4
(3) add and subtract x₀² = 16 (add zero to the equation)
We now have
x² - 8 x + 16 + y² - 2 y + 1 - 16 = 0
or
(x - 4)² + y² - 2 y + 1 - 16 = 0

Compare the y's for both equations (a) and (b):
(1) The coefficients of y² are the same; both are equal to one
(2) The coefficients of y are the same so -2 = -2 y₀ or y₀ = 1
(3) y₀² (= 1) is there in the equation
We now have
(x - 4)² + y² - 2 y + 1 - 16 = 0
or
(x - 4)² + (y - 1)² - 16 = 0
Moving the 16 to the other side and noting that 4² = 16, i.e. the square root of 16 is 4, we have
(x - 4)² + (y - 1)² = 4²
That is, a circle with center at (4, 1) and radius 4.

Now do the same with your equation.

 

[abel muroi]

You need to follow the steps. After getting a 1 in front of the x^2 and y^2 terms you take half of the coefficient of the x and y term and then square it and....

Sorry I am still having some trouble with this:???:

ok i re-did the problem again and this time i divided ALL sides

3x² +3y² + 6x - y = 0
x² + y² + 2x - 1/3y = 0 ((divided all sides by 3))
(x² +2x + 1) + (y² - 1/3y + 1/36) = 0 + 1 + 1/36 ((i multiplied 2x by 1/2 which got me 1 and also multiplied -1/3y with 1/2 which got me 1/36))
(x² + 1) + (y² + 1/36) = 37/36
(x + 1)² + (y + 1/6)² = 37/36

center = (1, 1/6) ((im not sure if these points have to be negative))
radius = 37/36

did i do the problem correctly this time?

 

[Ishuda]

Sorry I am still having some trouble with this:???:

ok i re-did the problem again and this time i divided ALL sides

3x² +3y² + 6x - y = 0
x² + y² + 2x - 1/3y = 0 ((divided all sides by 3))
(x² +2x + 1) + (y² - 1/3y + 1/36) = 0 + 1 + 1/36 ((i multiplied 2x by 1/2 which got me 1 and also multiplied -1/3y with 1/2 which got me 1/36))
(x² + 1) + (y² + 1/36) = 37/36
(x + 1)² + (y + 1/6)² = 37/36

center = (1, 1/6) ((im not sure if these points have to be negative))
radius = 37/36

did i do the problem correctly this time?

Almost: Note that the equation is
$\displaystyle (x - x_0)^2 + (y - y_0)^2 = r^2$
Thus
-x₀ = + 1
-y₀ = 1/6
r² = (37/36)

 

[abel muroi]

Almost: Note that the equation is
$\displaystyle (x - x_0)^2 + (y - y_0)^2 = r^2$
Thus
-x₀ = + 1
-y₀ = 1/6
r² = (37/36)

ohhhh i understand!

so since the equation has a negative inside the parenthesis, do i have to multiply THAT negative with 1 and 1/6? ((edit sorry i meant 1 and 1/16)

so the center becomes (-1, -1/6) ?

 

[Ishuda]

ohhhh i understand!

so since the equation has a negative inside the parenthesis, do i have to multiply THAT negative with 1 and 1/6? ((edit sorry i meant 1 and 1/16)

so the center becomes (-1, -1/6) ?

Yes
EDIT: For the equation as stated
(x + 1)² + (y + 1/6)² = 37/36
However, as Jomo points out below, that is not the proper solution for the original question.

 

[Deleted member 4993]

and the radius r = √(37)/6

 

[Steven G]

Sorry I am still having some trouble with this:???:

ok i re-did the problem again and this time i divided ALL sides

3x² +3y² + 6x - y = 0
x² + y² + 2x - 1/3y = 0 ((divided all sides by 3))
(x² +2x + 1) + (y² - 1/3y + 1/36) = 0 + 1 + 1/36 ((i multiplied 2x by 1/2 which got me 1 and also multiplied -1/3y with 1/2 which got me 1/36))
(x² + 1) + (y² + 1/36) = 37/36 What is this? Why does it follow from the line above and why does the next line follow from this? All, not some, not most but all equals signs must be valid
(x + 1)² + (y + 1/6)² = 37/36

center = (1, 1/6) ((im not sure if these points have to be negative)) To get the x-value of the center you solve x+1 =0. For the y value of the center you solve y+1/60
radius = 37/36 As you were told before the equation (x-x0)^2 + (y-y0)^2 = r^2. So 37/16 =r^2, not r

did i do the problem correctly this time?

my comments are above

 

[abel muroi]

my comments are above

wait.. I thought the radius was 37/6. or was it 37/16?

(x² + 1) + (y² + 1/36) = 37/36 What is this? Why does it follow from the line above and why does the next line follow from this? All, not some, not most but all equals signs must be valid

I added 1 and 1/36 which got me 37/36, then i square rooted 1/36 which got me 1/6

 

[Ishuda]

I'll add to the confusion: center is not (-1, -1/6), but is (-1, 1/6) :shock:

Ah, yes. I corrected my post after Jomo's post, see above(?)/back(?)/hopefully somewhere.

 

[Ishuda]

Huh? I was commenting on Abel's solution...what's you have to do with this?

Well Denis, by now you should know that I'm always involved. I just might not make a point of it. [I had made a post earlier where I had said Yes in answer to the question "so the center becomes (-1, -1/6) ?" which was correct if the equation for the circle had actually been as incorrectly given but was wrong in the context of the proper solution to the original problem. So, although I was right (again) I was actually wrong when I though I was right. This just shows to go that I am not always right (sometimes I'm left).

 

[Ishuda]

Are you a politician or a lawyer?

One could answer that question in the affirmative but then the correct answer would, of course, depend on the definition of lawyer and/or politician used {one could be both as well as just one of the choices and, of course, one could also be neither [not all at the same time though if one were an ordinary human (depending, of course, on one's definition of human)]}. How so some ever, I have never passed a bar (I've always gone in if it open and don't go to places where there are bars after they are closed) nor been elected to a public office {although, come to think of it, I have been elected to a private office after I became sufficiently senior [I am, BTW, quite a bit older than the average age of the population if quite a bit is described as at most 20 (and maybe even 25) years]}.

 

[Ishuda]

AHHHHHH....I get it: you're not either...but you're drunk :cool:

Not yet.