To find the center and radius of the circle given by the equation:
x^2 + y^2 -12x + 2y + 33 = 0
We need to rewrite the equation in standard form, which is:
(x - h)^2 + (y - k)^2 = r^2
where (h, k) is the center of the circle and r is the radius.
To do this, we need to complete the square for both x and y terms by adding and subtracting appropriate terms inside the parentheses.
(x^2 - 12x) + (y^2 + 2y) + 33 = 0
To complete the square for x terms, we need to add (12/2)^2 = 36 inside the first set of parentheses and subtract it from the equation:
(x^2 - 12x + 36) + (y^2 + 2y) + 33 - 36 = 0
(x - 6)^2 + (y^2 + 2y) - 3 = 0
To complete the square for y terms, we need to add (2/2)^2 = 1 inside the second set of parentheses and subtract it from the equation:
(x - 6)^2 + (y^2 + 2y + 1) - 4 = 0
(x - 6)^2 + (y + 1)^2 = 4
Now we can see that the equation is in standard form, where (h, k) = (6, -1) is the center of the circle, and r^2 = 4, so the radius is r = 2.
Therefore, the center of the circle is (6, -1) and the radius is 2.