To express [tex]\(x^2 - 2x + y^2 + 6y - 4 = 0\)[/tex] in standard form, we need to complete the square for both the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms. Follow these steps:
1. Group the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms:
[tex]\[
(x^2 - 2x) + (y^2 + 6y) - 4 = 0
\][/tex]
2. Move the constant term to the right side:
[tex]\[
(x^2 - 2x) + (y^2 + 6y) = 4
\][/tex]
3. Complete the square for the [tex]\(x\)[/tex] terms:
- Take the coefficient of [tex]\(x\)[/tex], which is [tex]\(-2\)[/tex], halve it to get [tex]\(-1\)[/tex], and square it to get [tex]\(1\)[/tex].
- Add and subtract this square inside the parentheses:
[tex]\[
x^2 - 2x = (x - 1)^2 - 1
\][/tex]
4. Complete the square for the [tex]\(y\)[/tex] terms:
- Take the coefficient of [tex]\(y\)[/tex], which is [tex]\(6\)[/tex], halve it to get [tex]\(3\)[/tex], and square it to get [tex]\(9\)[/tex].
- Add and subtract this square inside the parentheses:
[tex]\[
y^2 + 6y = (y + 3)^2 - 9
\][/tex]
5. Substitute these completed squares back into the equation:
[tex]\[
(x - 1)^2 - 1 + (y + 3)^2 - 9 = 4
\][/tex]
6. Combine the constants on the right side:
[tex]\[
(x - 1)^2 + (y + 3)^2 - 10 = 4
\][/tex]
7. Add [tex]\(10\)[/tex] to both sides to balance the equation:
[tex]\[
(x - 1)^2 + (y + 3)^2 = 14
\][/tex]
Therefore, the equation of the circle in standard form is:
[tex]\[
(x - 1)^2 + (y + 3)^2 = 14
\][/tex]
The correct answer is:
[tex]\[
\boxed{C. (x-1)^2+(y+3)^2=14}
\][/tex]
