Question

Mrs. Culland is finding the center of a circle whose equation

is x2 + y2 + 6x + 4y - 3 = 0 by completing the square. Her
work is shown.
x2 + y2 + 6x + 4y – 3 = 0
x2 + 6x + y2 + 4y - 3 = 0
(x2 + 6x) + (y2 + 4y) = 3
(x2 + 6x + 9) + (x2 + 4y + 4) = 3 + 9 + 4

Answer 1

Answer: ITS D !! ON EDGE

Explanation:

Answer 2

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

Explanation:

we know that

The equation of a circle in standard form is equal to

`(x-h)^(2)+(y-k)^(2)=r^(2)`

where

(h,k) is the center

r is the radius

In this problem we have

`x^(2) +y^(2)+6x+4y-3=0`

Completing the square

Group terms that contain the same variable, and move the constant to the opposite side of the equation

`(x^(2)+6x) +(y^(2)+4y)=3`

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

`(x^(2)+6x+9) +(y^(2)+4y+4)=3+9+4`

`(x^(2)+6x+9) +(y^(2)+4y+4)=16`

Rewrite as perfect squares

`(x+3)^(2) +(y+2)^(2)=16`

therefore

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