Question

mrs. culland is finding the center of a circle whose equation is $x^{2}+y^{2}+6x + 4y - 3 = 0$ by completing the square. her work is shown.$x^{2}+y^{2}+6x + 4y - 3 = 0$$x^{2}+6x + y^{2}+4y - 3 = 0$$(x^{2}+6x)+(y^{2}+4y)=3$$(x^{2}+6x + 9)+(y^{2}+4y + 4)=3 + 9 + 4$which completes the work correctly?$\bigcirc$ $(x - 3)^{2}+(y - 2)^{2}=4^{2}$, so the center is $(3, 2)$.$\bigcirc$ $(x + 3)^{2}+(y + 2)^{2}=4^{2}$, so the center is $(3, 2)$.$\bigcirc$ $(x - 3)^{2}+(y - 2)^{2}=4^{2}$, so the center is $(-3, -2)$.$\bigcirc$ $(x + 3)^{2}+(y + 2)^{2}=4^{2}$, so the center is $(-3, -2)$.

Explanation:

Step1: Simplify right-hand side

$3 + 9 + 4 = 16 = 4^2$

Step2: Rewrite left as perfect squares

$(x+3)^2 + (y+2)^2 = 4^2$

Step3: Identify circle center

For $(x-h)^2+(y-k)^2=r^2$, center is $(h,k)$. Here $h=-3, k=-2$.

Answer:

$(x + 3)^2 + (y + 2)^2 = 4^2$, so the center is $(-3, -2)$.