To find the center of the circle given the equation [tex]\(x^2 + y^2 + 4x - 8y + 11 = 0\)[/tex], we need to rewrite the equation in the standard form of a circle's equation, which is [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex]. Here, [tex]\((h, k)\)[/tex] is the center of the circle, and [tex]\(r\)[/tex] is the radius.
We will complete the square for both the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms in the equation.
1. Start with the given equation:
[tex]\[
x^2 + y^2 + 4x - 8y + 11 = 0
\][/tex]
2. Group the [tex]\(x\)[/tex] terms and the [tex]\(y\)[/tex] terms together:
[tex]\[
(x^2 + 4x) + (y^2 - 8y) + 11 = 0
\][/tex]
3. Complete the square for the [tex]\(x\)[/tex] terms:
- Take the coefficient of [tex]\(x\)[/tex], which is 4, divide it by 2 and square the result:
[tex]\[
\left(\frac{4}{2}\right)^2 = 4
\][/tex]
- Add and subtract this square inside the equation:
[tex]\[
(x^2 + 4x + 4 - 4) + (y^2 - 8y) + 11 = 0
\][/tex]
- Simplify the expression as a perfect square:
[tex]\[
(x + 2)^2 - 4 + (y^2 - 8y) + 11 = 0
\][/tex]
4. Complete the square for the [tex]\(y\)[/tex] terms:
- Take the coefficient of [tex]\(y\)[/tex], which is -8, divide it by 2 and square the result:
[tex]\[
\left(\frac{-8}{2}\right)^2 = 16
\][/tex]
- Add and subtract this square inside the equation:
[tex]\[
(x + 2)^2 - 4 + (y^2 - 8y + 16 - 16) + 11 = 0
\][/tex]
- Simplify the expression as a perfect square:
[tex]\[
(x + 2)^2 - 4 + (y - 4)^2 - 16 + 11 = 0
\][/tex]
5. Combine constant terms:
[tex]\[
(x + 2)^2 + (y - 4)^2 - 9 = 0
\][/tex]
6. Move the constant term to the other side of the equation to get the standard form:
[tex]\[
(x + 2)^2 + (y - 4)^2 = 9
\][/tex]
From the standard form [tex]\((x + 2)^2 + (y - 4)^2 = 9\)[/tex], we see that the center of the circle is [tex]\((-2, 4)\)[/tex].
Therefore, the center of the circle is [tex]\(\boxed{(-2, 4)}\)[/tex].
