Sure, let's solve the problem step by step.
### Part A: Complete the Square to Rewrite the Equation in Standard Form
Given the equation:
[tex]\[ x^2 + 4x + y^2 - 6y = -4 \][/tex]
We need to complete the square for both [tex]\( x \)[/tex]-terms and [tex]\( y \)[/tex]-terms.
#### Step 1: Completing the Square for the [tex]\( x \)[/tex]-Terms
The expression involving [tex]\( x \)[/tex] is [tex]\( x^2 + 4x \)[/tex].
1. Take half of the coefficient of [tex]\( x \)[/tex] (which is 4), and square it:
[tex]\[ \left( \frac{4}{2} \right)^2 = 2^2 = 4 \][/tex]
2. Add and then subtract this square inside the equation:
[tex]\[ x^2 + 4x = (x^2 + 4x + 4) - 4 = (x + 2)^2 - 4 \][/tex]
#### Step 2: Completing the Square for the [tex]\( y \)[/tex]-Terms
The expression involving [tex]\( y \)[/tex] is [tex]\( y^2 - 6y \)[/tex].
1. Take half of the coefficient of [tex]\( y \)[/tex] (which is -6), and square it:
[tex]\[ \left( \frac{-6}{2} \right)^2 = (-3)^2 = 9 \][/tex]
2. Add and then subtract this square inside the equation:
[tex]\[ y^2 - 6y = (y^2 - 6y + 9) - 9 = (y - 3)^2 - 9 \][/tex]
#### Step 3: Rewriting the Original Equation
Now, substitute the completed squares back into the equation:
[tex]\[
x^2 + 4x + y^2 - 6y = -4
\][/tex]
[tex]\[ (x + 2)^2 - 4 + (y - 3)^2 - 9 = -4 \][/tex]
Combine like terms:
[tex]\[
(x + 2)^2 + (y - 3)^2 - 13 = -4
\][/tex]
Add 13 to both sides to get the equation in standard form:
[tex]\[
(x + 2)^2 + (y - 3)^2 = 9
\][/tex]
This is the standard form of a circle equation:
[tex]\[ (x - h)^2 + (y - k)^2 = r^2 \][/tex]
### Part B: Identify the Center and Radius of the Circle
From the standard form equation [tex]\( (x + 2)^2 + (y - 3)^2 = 9 \)[/tex]:
1. The center [tex]\((h, k)\)[/tex] of the circle is found by looking at the transformations:
- [tex]\( (x + 2) \)[/tex] indicates a horizontal shift of -2
- [tex]\( (y - 3) \)[/tex] indicates a vertical shift of 3
Thus, the center is:
[tex]\[
(h, k) = (-2, 3)
\][/tex]
2. The radius [tex]\( r \)[/tex]:
- The equation [tex]\( (x + 2)^2 + (y - 3)^2 = 9 \)[/tex] has [tex]\( r^2 = 9 \)[/tex]
- Therefore, the radius is:
[tex]\[
r = \sqrt{9} = 3
\][/tex]
So, the center of the circle is [tex]\((-2, 3)\)[/tex] and the radius is [tex]\(3\)[/tex].
