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: Group $x$ and $y$ terms

We have $(x^{2}+4x)+(y^{2}-6y)=36$.

Step3: Complete the square for $x$ - terms

For the $x$ - terms $x^{2}+4x$, we add $(\frac{4}{2})^2 = 4$ to both sides of the equation. For the $y$ - terms $y^{2}-6y$, we add $(\frac{-6}{2})^2=9$ to both sides of the equation. So, $(x^{2}+4x + 4)+(y^{2}-6y + 9)=36 + 4+9$.

Step4: Rewrite in standard form

$(x + 2)^2+(y - 3)^2=49$, which is in the form $(x - a)^2+(y - b)^2=r^2$ with $a=-2$, $b = 3$, and $r = 7$.

Answer:

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