To determine the center of the circle given by the equation [tex]\( x^2 + y^2 + 4x - 8y + 11 = 0 \)[/tex], we will rewrite this equation in the standard form of a circle’s equation, which is [tex]\( (x - h)^2 + (y - k)^2 = r^2 \)[/tex]. This standard form allows us to identify the center [tex]\((h, k)\)[/tex] directly.
### Steps to Rewrite the Equation
1. Group the [tex]\(x\)[/tex] terms and the [tex]\(y\)[/tex] terms:
[tex]\[
x^2 + 4x + y^2 - 8y = -11
\][/tex]
2. Complete the square for the [tex]\(x\)[/tex] terms.
- For [tex]\(x^2 + 4x\)[/tex], take half the coefficient of [tex]\(x\)[/tex] (which is 4), square it, and add inside the square while balancing it outside:
[tex]\[
x^2 + 4x \implies x^2 + 4x + 4 - 4 = (x + 2)^2 - 4
\][/tex]
3. Complete the square for the [tex]\(y\)[/tex] terms.
- For [tex]\(y^2 - 8y\)[/tex], take half the coefficient of [tex]\(y\)[/tex] (which is -8), square it, and add inside the square while balancing it outside:
[tex]\[
y^2 - 8y \implies y^2 - 8y + 16 - 16 = (y - 4)^2 - 16
\][/tex]
4. Combine these completed squares back into the equation.
[tex]\[
(x + 2)^2 - 4 + (y - 4)^2 - 16 = -11
\][/tex]
5. Simplify by combining constants:
[tex]\[
(x + 2)^2 + (y - 4)^2 - 20 = -11 \implies (x + 2)^2 + (y - 4)^2 = 9
\][/tex]
The equation [tex]\((x + 2)^2 + (y - 4)^2 = 9\)[/tex] is now in the standard form of a circle, where [tex]\((h, k)\)[/tex] is the center of the circle, and [tex]\(r^2 = 9\)[/tex], so the radius is 3.
From this standard form, we can identify the center of the circle [tex]\((h, k)\)[/tex]:
### Center of the Circle
[tex]\[
(h, k) = (-2, 4)
\][/tex]
Thus, the center of the circle given by the equation [tex]\( x^2 + y^2 + 4x - 8y + 11 = 0 \)[/tex] is [tex]\(\boxed{(-2, 4)}\)[/tex].
