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

∵ 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)

10% of increase is the final answer! : )