Mrs. Culland is finding the center of a circle whose equationis x2 + y2 + 6x + 4y - 3 = 0 by completing the square. Herwork is shown.x2 + y2 + 6x + 4y – 3 = 0x2 + 6x + y2 + 4y - 3 = 0(x2 + 6x) + (y2 + 4y) = 3(x2 + 6x + 9) + (x2 + 4y + 4) = 3 + 9 + 4

Answer:The center of the circle is (-3,-2)Step-by-step explanation:we know thatThe equation of a circle in standard form is equal to[tex](x-h)^{2}+(y-k)^{2}=r^{2}[/tex]where(h,k) is the centerr is the radiusIn this problem we have[tex]x^{2} +y^{2}+6x+4y-3=0[/tex]Completing the squareGroup terms that contain the same variable, and move the constant to the opposite side of the equation[tex](x^{2}+6x) +(y^{2}+4y)=3[/tex]Complete the square twice. Remember to balance the equation by adding the same constants to each side.[tex](x^{2}+6x+9) +(y^{2}+4y+4)=3+9+4[/tex][tex](x^{2}+6x+9) +(y^{2}+4y+4)=16[/tex]Rewrite as perfect squares[tex](x+3)^{2} +(y+2)^{2}=16[/tex]thereforeThe center of the circle is (-3,-2)