Find the center and radius of the circular equation x^2 + y^2 - 12y + 20 = 0 ?

To find the center and radius of the circle given by the equation x^2 + y^2 - 12y + 20 = 0, we first need to rewrite the equation in standard form.

Complete the square for the y terms:
y^2 - 12y = y^2 - 12y + 36 - 36 = (y - 6)^2 - 36

Now rewrite the equation:
x^2 + (y - 6)^2 - 36 + 20 = 0
x^2 + (y - 6)^2 - 16 = 0

Now the equation is in standard form:
(x - 0)^2 + (y - 6)^2 = 4^2

Comparing this to the standard form of a circle equation:
(x - h)^2 + (y - k)^2 = r^2

We can see that the center of the circle is at (0, 6) and the radius is 4.

Complete the square for both x and y:

x^2 + y^2 - 12y + 20 = 0
x^2 + y^2 - 12y + 36 - 36 + 20 = 0
(x^2 - 36) + (y^2 - 12y + 36) = 16
(x - 6)^2 + (y - 6)^2 = 4^2

So the center is (6, 6) and the radius is 4.