What is the center of the circle described by the equation

x^2+4x+y^2-6y=12

(4, - 6)

(-4, 6)

(-2, 3)

(2, - 3)

The center of the circle is (-2, 3) ⇒ 3rd answer

Step-by-step explanation:

The equation of a circle is (x - h) ² + (y - k) ² = r², where

(h, k) are the coordinates of its center r is the radius of it

∵ The equation of the circle is x² + 4x + y² - 6y = 12

- Lets make a completing square for x² + 4x

∵ x² = (x) (x)

∵ 4x : 2 = 2x

- That means the second term of the bracket (x + ...) ² is 2

∴ The bracket is (x + 2)

∵ (x + 2) ² = x² + 4x + 4

∴ We must add 4 and subtract 4 in the equation of the circle

∴ (x² + 4x + 4) - 4 + y² - 6y = 12

Lets make a completing square for y² - 6y

∵ y² = (y) (y)

∵ - 6y : 2 = - 3y

- That means the second term of the bracket (y + ...) is - 3

∴ The bracket is (y - 3)

∵ (y - 3) ² = y² - 6y + 9

∴ We must add 9 and subtract 9 in the equation of the circle

∴ (x² + 4x + 4) - 4 + (y² - 6y + 9) - 9 = 12

Now lets simplify the equation

∵ (x + 2) ² + (y - 3) ² - 13 = 12

- Add 13 to both sides

∴ (x + 2) ² + (y - 3) ² = 25

- Compare it with the form of the equation of the circle to

find h and k

∵ (x - h) ² + (y - k) ² = r²

∴ h = - 2 and k = 3

The center of the circle is (-2, 3)