Question

Find the center and the radius of the circle with the equation:

x^2 + 6x + y^2 + 4y + 12 = 0
a.

center: (3, 2)
radius: 1

c.

center: (6, 4)
radius: 12
b.

center: (-3, -2)
radius: 1

d.

center: (-6, -4)
radius: 12

Answer 1

Center is (-3, -2) and radius is 1

Explanation:

• Step 1: Given equation of the circle is x² + 6x + y² + 4y + 12 = 0. Standard form is x² + y² + 2gx + 2fy + c = 0. Center is (-g, -f) and radius is √g² + f² - c. Find g, f and c.

By comparing the 2 equations,

⇒ 2gx = 6x

⇒ 2g = 6

∴ g = 6/2 = 3

⇒ 2fy = 4y

⇒ 2f = 4

∴ f = 4/2 = 2

⇒ c = 12

• Step 2: Find center and radius.

Center = (-g, -f) = (-3, -2)

Radius = √g² + f² - c

= √3² + 2² - 12 = √9 + 4 - 12

= 1