To find the center of the circle given by the equation [tex]\( x^2 + y^2 - 4x + 2y - 11 = 0 \)[/tex], we will rewrite it into the standard form of a circle's equation, [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex]. The steps are as follows:
1. Group the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms together:
[tex]\[
x^2 - 4x + y^2 + 2y = 11
\][/tex]
We achieved this by moving the constant term [tex]\(-11\)[/tex] to the right side of the equation.
2. Complete the square for the [tex]\(x\)[/tex]-terms and [tex]\(y\)[/tex]-terms:
- For the [tex]\(x\)[/tex]-terms [tex]\(x^2 - 4x\)[/tex]:
- Take half of the coefficient of [tex]\(x\)[/tex], which is [tex]\(-4\)[/tex], divide it by 2 to get [tex]\(-2\)[/tex], and then square it to get 4. Add and subtract 4 inside the equation.
- [tex]\((x^2 - 4x + 4) - 4\)[/tex]
- Rewrite as [tex]\((x - 2)^2 - 4\)[/tex]
- For the [tex]\(y\)[/tex]-terms [tex]\(y^2 + 2y\)[/tex]:
- Take half of the coefficient of [tex]\(y\)[/tex], which is 2, divide it by 2 to get 1, and then square it to get 1. Add and subtract 1 inside the equation.
- [tex]\((y^2 + 2y + 1) - 1\)[/tex]
- Rewrite as [tex]\((y + 1)^2 - 1\)[/tex]
So the equation with completed squares is:
[tex]\[
(x - 2)^2 - 4 + (y + 1)^2 - 1 = 11
\][/tex]
3. Combine the constants on the right side of the equation:
[tex]\[
(x - 2)^2 + (y + 1)^2 - 5 = 11
\][/tex]
[tex]\[
(x - 2)^2 + (y + 1)^2 = 16
\][/tex]
Now, we have the equation in the standard form [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex] where [tex]\(h = 2\)[/tex], [tex]\(k = -1\)[/tex], and [tex]\(r^2 = 16\)[/tex].
Thus, the center of the circle [tex]\((h, k)\)[/tex] is [tex]\((2, -1)\)[/tex].
So the correct answer is:
D. [tex]\((2, -1)\)[/tex]
