Question

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

Answer 1

Answer:

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

Step-by-step 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)