The question is; identify the center and the radius of the circle in this equation.

3x^2 - √ 3x + 3y^2 - y + 1/12 = 0

I ended up getting

C:(√ 3/6, 1/6)

and

R: √ 3/6

To identify the center and radius of the circle in the equation 3x^2 - √3x + 3y^2 - y + 1/12 = 0, we'll need to rearrange the equation into the standard form of the equation of a circle, which is (x - h)^2 + (y - k)^2 = r^2.

Let's start by completing the square for the x terms:
3x^2 - √3x + 3y^2 - y + 1/12 = 0
3(x^2 - (√3/3)x) + 3y^2 - y + 1/12 = 0
3(x^2 - (√3/3)x + (√3/6)^2) + 3y^2 - y + 1/12 - 3(√3/6)^2 = 0
3(x - √3/6)^2 + 3y^2 - y + 1/12 - 3/72 = 0
3(x - √3/6)^2 + 3y^2 - y + 1/12 - 1/24 = 0
3(x - √3/6)^2 + 3y^2 - y + 1/24 = 0

Now let's complete the square for the y terms:
3(x - √3/6)^2 + 3y^2 - y + 1/24 = 0
3(x - √3/6)^2 + 3(y^2 - (1/3)y) + 1/24 = 0
3(x - √3/6)^2 + 3(y^2 - (1/3)y + (1/6)^2) + 1/24 - 3(1/6)^2 = 0
3(x - √3/6)^2 + 3(y - 1/6)^2 + 1/24 - 1/12 = 0
3(x - √3/6)^2 + 3(y - 1/6)^2 + 1/24 - 2/24 = 0
3(x - √3/6)^2 + 3(y - 1/6)^2 - 1/24 = 0

Now we can see that the equation is in the standard form of the equation of a circle, where (h, k) is the center of the circle and r is the radius.

From the equation, we can identify that:
Center (h, k) = (√3/6, 1/6)
Radius r = √1/24 = √3/6

Therefore, the center of the circle is (√3/6, 1/6) and the radius is √3/6.

To identify the center and radius of a circle given its equation, we need to put the equation in a standard form of a circle, which is (x - h)^2 + (y - k)^2 = r^2.

Let's start by rearranging the given equation:

3x^2 - √3x + 3y^2 - y + 1/12 = 0

To make it easier to work with, let's multiply the whole equation by 12 to remove the fraction:

36x^2 - 12√3x + 36y^2 - 12y + 1 = 0

Now, let's complete the square for the x and y terms.

For the x terms:
We can rewrite it as (6x - (√3/2))^2 = (6x)^2 - 2 * 6x * (√3/2) + (√3/2)^2 = 36x^2 - 6√3x + 3/4

For the y terms:
We can rewrite it as (6y - 1/2)^2 = (6y)^2 - 2 * 6y * (1/2) + (1/2)^2 = 36y^2 - 6y + 1/4

Plugging these square completions back into the equation, we get:

(6x - (√3/2))^2 + (6y - 1/2)^2 = 36x^2 - 6√3x + 3/4 + 36y^2 - 6y + 1/4

Combining like terms, we have:

(6x - (√3/2))^2 + (6y - 1/2)^2 = 36x^2 + 36y^2 - 6√3x - 6y + 1

Comparing it to the standard form of a circle, we can see that the center of the circle is the point (h, k), which is equal to (√3/6, 1/6).

The radius of the circle, denoted as r, is equal to the square root of the constant term on the right side of the equation. In this case, the radius is √(36x^2 + 36y^2 - 6√3x - 6y + 1), which simplifies to √3/6.

Therefore, the center of the circle is C: (√3/6, 1/6), and the radius is R: √3/6.