Find the center and the radius of the circle with the equation x^2+6x+y^2+4y+12=0

(x + 3)² + (y+ 2)² = 1²

Step-by-step explanation:

The standard form for the equation of a circle with centre (a, b) and radius r is

(x - a)² + (y - b)² = r²

x² + 6x + y² + 4y + 12 =  0

Step 1. Subtract the constant from each side

x² + 6x +           y² + 4y        = -12

Step 2. Complete the squares for x and y

Square half the coefficients of x and y

(x² + 6x + 3²) + (y² + 4y + 2²) = -12 + 3² + 2²

Step 3. Write the lhs as squares of binomials

(x + 3)² + (y+ 2)² = -12 + 3² + 2²

Step 4. Convert the rhs to a square

(x + 3)² + (y+ 2)² = -12 + 9 + 4

(x + 3)² + (y+ 2)² = 1

(x + 3)² + (y+ 2)² = 1²

The graph below shows that this is the equation for a circle with

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

answer: 3.26019516

step-by-step explanation: