Final answer:
To rewrite the equation 3x^2 + 3y^2 + 6x - y = 0 in standard form, we group the x terms together and the y terms together, complete the square for x and y, and simplify the equation. The standard form of the equation is (x + 3)^2 + (3y - 1/2)^2 = 81/4. The center of the circle is (-3, 1/6) and the radius is 9/2.
Step-by-step explanation:
To rewrite the equation 3x^2 + 3y^2 + 6x - y = 0 in standard form, we need to complete the square for both the x and y terms. First, let's group the x terms together and the y terms together:
(3x^2 + 6x) + (3y^2 - y) = 0
Now, let's complete the square for x by adding half of the coefficient of x (which is 3) squared to both sides of the equation:
(3x^2 + 6x + 3^2) + (3y^2 - y) = 3^2
Simplifying this gives us:
(x + 3)^2 + (3y^2 - y) = 9
Next, let's complete the square for y by adding half of the coefficient of y (which is -1) squared to both sides of the equation:
(x + 3)^2 + (3y^2 - y + (-1/2)^2) = 9 + (-1/2)^2
Simplifying this gives us:
(x + 3)^2 + (3y - 1/2)^2 = 81/4
Now the equation is in standard form, (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center of the circle and r is the radius. Therefore, the center of the circle is (-3, 1/6) and the radius is √(81/4), which simplifies to 9/2.