Question

consider a circle whose equation is (x^{2}+y^{2}+4x - 6y-36 = 0). which statements are true? check all that apply. to begin converting the equation to standard form, subtract 36 from both sides. to complete the square for the x terms, add 4 to both sides. the center of the circle is at (-2, 3). the center of the circle is at (4, -6). the radius of the circle is 6 units. the radius of the circle is 49 units.

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. Given $x^{2}+y^{2}+4x - 6y-36 = 0$.

Step2: Rearrange terms

Group the $x$ - terms and $y$ - terms: $(x^{2}+4x)+(y^{2}-6y)=36$.

Step3: Complete the square for $x$

For the $x$ - terms $x^{2}+4x$, we know that $(x + m)^2=x^{2}+2mx+m^{2}$. Here $2m = 4$, so $m = 2$ and we add $m^{2}=4$ to both sides of the equation: $(x^{2}+4x + 4)+(y^{2}-6y)=36 + 4$.

Step4: Complete the square for $y$

For the $y$ - terms $y^{2}-6y$, since $2n=-6$, $n=-3$ and $n^{2}=9$. Add 9 to both sides: $(x^{2}+4x + 4)+(y^{2}-6y + 9)=36 + 4+9$.

Step5: Write in standard form

$(x + 2)^2+(y - 3)^2=49$, which is $(x-(-2))^2+(y - 3)^2=7^2$. The center of the circle is $(-2,3)$ and the radius $r = 7$.

Answer:

To complete the square for the $x$ terms, add 4 to both sides.
The center of the circle is at $(-2,3)$.