To determine the coordinates of the center of the circle given by the equation [tex]\( x^2 + 2x + y^2 - 4y = 12 \)[/tex], we need to rewrite it in the standard form of a circle's equation, which is [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex], where [tex]\((h, k)\)[/tex] is the center of the circle and [tex]\(r\)[/tex] is the radius.
Here are the steps to find the center of the circle by completing the square for both [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
1. Complete the square for [tex]\(x\)[/tex]:
The quadratic expression involving [tex]\(x\)[/tex] is [tex]\(x^2 + 2x\)[/tex].
- To complete the square, take the coefficient of [tex]\(x\)[/tex], which is 2, divide it by 2 to get 1, and then square it to get 1.
- Rewrite [tex]\(x^2 + 2x\)[/tex] as [tex]\((x + 1)^2 - 1\)[/tex].
2. Complete the square for [tex]\(y\)[/tex]:
The quadratic expression involving [tex]\(y\)[/tex] is [tex]\(y^2 - 4y\)[/tex].
- To complete the square, take the coefficient of [tex]\(y\)[/tex], which is -4, divide it by 2 to get -2, and then square it to get 4.
- Rewrite [tex]\(y^2 - 4y\)[/tex] as [tex]\((y - 2)^2 - 4\)[/tex].
Now substitute these completed square forms back into the original equation:
[tex]\[
x^2 + 2x + y^2 - 4y = 12
\][/tex]
becomes
[tex]\[
(x + 1)^2 - 1 + (y - 2)^2 - 4 = 12
\][/tex]
Simplify this equation:
[tex]\[
(x + 1)^2 + (y - 2)^2 - 5 = 12
\][/tex]
Add 5 to both sides to isolate the squared terms:
[tex]\[
(x + 1)^2 + (y - 2)^2 = 17
\][/tex]
This is now in the standard form [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex].
From the rewritten equation, we can identify the center [tex]\((h, k)\)[/tex] of the circle:
[tex]\[
h = -1
\][/tex]
[tex]\[
k = 2
\][/tex]
Thus, the coordinates of the center of the circle are [tex]\((-1, 2)\)[/tex].
