Question

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

Explanation:

Step1: Rearrange constant term

Move -36 to the right side:
$x^2 + 4x + y^2 - 6y = 36$

Step2: Complete square for x terms

Take half of 4, square it ($(\frac{4}{2})^2=4$), add to both sides:
$x^2 + 4x + 4 + y^2 - 6y = 36 + 4$

Step3: Complete square for y terms

Take half of -6, square it ($(\frac{-6}{2})^2=9$), add to both sides:
$(x+2)^2 + y^2 - 6y + 9 = 40 + 9$

Step4: Write standard circle form

Simplify to standard form $(x-h)^2+(y-k)^2=r^2$:
$(x+2)^2 + (y-3)^2 = 49$

Step5: Identify center and radius

Center $(h,k)=(-2,3)$, radius $r=\sqrt{49}=7$

Answer:

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