To find the circle and radius of equation x2 + y2 +12x +6y +20=0, we need to complete the square for both x and y terms.
First, we can simplify the equation by rearranging the constant term to the right-hand side:
x^2 + y^2 + 12x + 6y = -20
Next, we can complete the square for the x-terms by adding (12/2)^2 = 36 to both sides of the equation:
x^2 + 12x + 36 + y^2 + 6y = -20 + 36
Simplifying further:
(x + 6)^2 + (y + 3)^2 = 13
This is now in the standard form of a circle: (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center of the circle and r is the radius.
So the center of the circle is (-6, -3) and the radius is the square root of 13. Therefore, the equation x^2 + y^2 + 12x + 6y + 20 = 0 represents a circle with center (-6, -3) and radius sqrt(13).