mrs. culland is finding the center of a circle whose equation is $x^{2}+y^{2}+6x + 4y-3 = 0$ by completing…

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.\n$x^{2}+y^{2}+6x + 4y-3 = 0$\n$x^{2}+6x + y^{2}+4y-3 = 0$\n$(x^{2}+6x)+(y^{2}+4y)=3$\n$(x^{2}+6x + 9)+(y^{2}+4y + 4)=3 + 9+4$\nwhich completes the work correctly?\n$(x - 3)^{2}+(y - 2)^{2}=4^{2}$, so the center is $(3,2)$.\n$(x + 3)^{2}+(y + 2)^{2}=4^{2}$, so the center is $(3,2)$.\n$(x - 3)^{2}+(y - 2)^{2}=4^{2}$, so the center is $(-3,-2)$.\n$(x + 3)^{2}+(y + 2)^{2}=4^{2}$, so the center is $(-3,-2)$.

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.\n$x^{2}+y^{2}+6x + 4y-3 = 0$\n$x^{2}+6x + y^{2}+4y-3 = 0$\n$(x^{2}+6x)+(y^{2}+4y)=3$\n$(x^{2}+6x + 9)+(y^{2}+4y + 4)=3 + 9+4$\nwhich completes the work correctly?\n$(x - 3)^{2}+(y - 2)^{2}=4^{2}$, so the center is $(3,2)$.\n$(x + 3)^{2}+(y + 2)^{2}=4^{2}$, so the center is $(3,2)$.\n$(x - 3)^{2}+(y - 2)^{2}=4^{2}$, so the center is $(-3,-2)$.\n$(x + 3)^{2}+(y + 2)^{2}=4^{2}$, so the center is $(-3,-2)$.

Answer

Answer:

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

Explanation:

Step1: Recall circle - standard form

The standard form of a circle equation is $(x - a)^2+(y - b)^2=r^2$, where $(a,b)$ is the center and $r$ is the radius.

Step2: Complete the square for $x$ - terms

For $x^{2}+6x$, we know that $(x + m)^2=x^{2}+2mx+m^{2}$. Here $2m = 6$, so $m = 3$ and $x^{2}+6x=(x + 3)^2-9$.

Step3: Complete the square for $y$ - terms

For $y^{2}+4y$, since $2n = 4$, then $n = 2$ and $y^{2}+4y=(y + 2)^2-4$.

Step4: Rewrite the given equation

The original equation $x^{2}+y^{2}+6x + 4y-3=0$ can be rewritten as $(x^{2}+6x)+(y^{2}+4y)=3$. Substituting the completed - square forms, we get $(x + 3)^2-9+(y + 2)^2-4=3$.

Step5: Simplify the equation

$(x + 3)^2+(y + 2)^2=3 + 9+4=16=4^2$.

Step6: Identify the center

Comparing with the standard form $(x - a)^2+(y - b)^2=r^2$, we have $a=-3$, $b=-2$. So the center of the circle is $(-3,-2)$.