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}](https://tex.z-dn.net/?f=%28x-h%29%5E%7B2%7D%2B%28y-k%29%5E%7B2%7D%3Dr%5E%7B2%7D)
where
(h,k) is the center
r is the radius
In this problem we have
![x^{2} +y^{2}+6x+4y-3=0](https://tex.z-dn.net/?f=x%5E%7B2%7D%20%2By%5E%7B2%7D%2B6x%2B4y-3%3D0)
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](https://tex.z-dn.net/?f=%28x%5E%7B2%7D%2B6x%29%20%2B%28y%5E%7B2%7D%2B4y%29%3D3)
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](https://tex.z-dn.net/?f=%28x%5E%7B2%7D%2B6x%2B9%29%20%2B%28y%5E%7B2%7D%2B4y%2B4%29%3D3%2B9%2B4)
![(x^{2}+6x+9) +(y^{2}+4y+4)=16](https://tex.z-dn.net/?f=%28x%5E%7B2%7D%2B6x%2B9%29%20%2B%28y%5E%7B2%7D%2B4y%2B4%29%3D16)
Rewrite as perfect squares
![(x+3)^{2} +(y+2)^{2}=16](https://tex.z-dn.net/?f=%28x%2B3%29%5E%7B2%7D%20%2B%28y%2B2%29%5E%7B2%7D%3D16)
therefore
The center of the circle is (-3,-2)