Find the center and the radius of the circle with the equation:
x2+6x+y2+4y+12=0

centre = (- 3, - 2) , radius = 1

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

Given

x² + 6x + y² + 4y + 12 = 0 ( subtract 12 from both sides )

x² + 6x + y² + 4y = - 12

Use the method of completing the square on the x and y terms

add ( half the coefficient of the x/ y terms )² to both sides

x² + 2(3)x + 9 + y² + 2(2)y + 4 = - 12 + 9 + 4

(x + 3)² + (y + 2)² = 1 ← in standard form

with centre = (- (- 3), - (- 2)) and r² = 1, that is

centre = (- 3, - 2) and radius = 1