### Part A: Complete the Square
To rewrite the given equation [tex]\(x^2 + 4x + y^2 - 6y = -4\)[/tex] in standard form, we will complete the square for the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms separately.
1. Group the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms:
[tex]\[
x^2 + 4x + y^2 - 6y = -4
\][/tex]
2. Complete the square for the [tex]\(x\)[/tex]-terms ([tex]\(x^2 + 4x\)[/tex]):
- Take the coefficient of [tex]\(x\)[/tex], which is 4, halve it to get 2, and then square it to get 4.
- Add and subtract 4 within the equation:
[tex]\[
x^2 + 4x = (x + 2)^2 - 4
\][/tex]
3. Complete the square for the [tex]\(y\)[/tex]-terms ([tex]\(y^2 - 6y\)[/tex]):
- Take the coefficient of [tex]\(y\)[/tex], which is -6, halve it to get -3, and then square it to get 9.
- Add and subtract 9 within the equation:
[tex]\[
y^2 - 6y = (y - 3)^2 - 9
\][/tex]
4. Substitute these into the original equation:
[tex]\[
(x + 2)^2 - 4 + (y - 3)^2 - 9 = -4
\][/tex]
5. Simplify the equation:
- Combine constants on the left side:
[tex]\[
(x + 2)^2 + (y - 3)^2 - 13 = -4
\][/tex]
- Add 13 to both sides to isolate the squared terms:
[tex]\[
(x + 2)^2 + (y - 3)^2 = 9
\][/tex]
Therefore, the equation in standard form is:
[tex]\[
(x + 2)^2 + (y - 3)^2 = 9
\][/tex]
### Part B: Identify the Center and Radius
From the standard form equation [tex]\((x + 2)^2 + (y - 3)^2 = 9\)[/tex], we can identify the geometry of the circle:
- Center: The circle has its center at [tex]\((h, k)\)[/tex], where [tex]\(h\)[/tex] and [tex]\(k\)[/tex] are taken from the expressions [tex]\((x + 2)^2\)[/tex] and [tex]\((y - 3)^2\)[/tex]. Thus, the center is:
[tex]\[
(-2, 3)
\][/tex]
- Radius: The radius [tex]\(r\)[/tex] can be determined from the right side of the equation, which is 9. Since the equation is in the form [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex]:
[tex]\[
r^2 = 9 \implies r = \sqrt{9} = 3
\][/tex]
Thus, the center of the circle is [tex]\((-2, 3)\)[/tex] and the radius is [tex]\(3\)[/tex].
