Answer :
### Part A: Complete the square to rewrite the equation in standard form
Step 1: Start with the given equation:
[tex]\[ x^2 + 2x + y^2 + 4y = 20 \][/tex]
Step 2: Group the [tex]\(x\)[/tex]-terms and [tex]\(y\)[/tex]-terms:
[tex]\[ (x^2 + 2x) + (y^2 + 4y) = 20 \][/tex]
Step 3: Complete the square for the [tex]\(x\)[/tex]-terms:
For the quadratic expression [tex]\(x^2 + 2x\)[/tex], take half of the coefficient of [tex]\(x\)[/tex] (which is 2), square it, and add and subtract this square inside the equation.
[tex]\[ x^2 + 2x = (x + 1)^2 - 1 \][/tex]
Step 4: Complete the square for the [tex]\(y\)[/tex]-terms:
Similarly, for [tex]\(y^2 + 4y\)[/tex], take half of the coefficient of [tex]\(y\)[/tex] (which is 4), square it, and add and subtract this square inside the equation.
[tex]\[ y^2 + 4y = (y + 2)^2 - 4 \][/tex]
Step 5: Rewrite the equation with the completed squares:
[tex]\[ (x + 1)^2 - 1 + (y + 2)^2 - 4 = 20 \][/tex]
Add the constants (-1 and -4) to both sides of the equation to balance it:
[tex]\[ (x + 1)^2 + (y + 2)^2 - 5 = 20 \][/tex]
[tex]\[ (x + 1)^2 + (y + 2)^2 = 25 \][/tex]
This is now in standard form:
[tex]\[ (x + 1)^2 + (y + 2)^2 = 25 \][/tex]
### Part B: Identify the center and radius of the circle
Step 1: Identify the center [tex]\((h, k)\)[/tex]:
From the standard form of the circle equation [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex]:
Our equation is [tex]\((x + 1)^2 + (y + 2)^2 = 25\)[/tex].
Compare this with [tex]\( (x - (-1))^2 + (y - (-2))^2 = 25 \)[/tex]:
The center [tex]\((h, k)\)[/tex] is:
[tex]\[ (-1, -2) \][/tex]
Step 2: Identify the radius [tex]\(r\)[/tex]:
The right side of the equation is [tex]\(r^2 = 25\)[/tex]. The radius [tex]\(r\)[/tex] is the square root of 25:
[tex]\[ r = \sqrt{25} = 5 \][/tex]
Summary:
- The center of the circle is [tex]\((-1, -2)\)[/tex].
- The radius of the circle is [tex]\(5\)[/tex].
Step 1: Start with the given equation:
[tex]\[ x^2 + 2x + y^2 + 4y = 20 \][/tex]
Step 2: Group the [tex]\(x\)[/tex]-terms and [tex]\(y\)[/tex]-terms:
[tex]\[ (x^2 + 2x) + (y^2 + 4y) = 20 \][/tex]
Step 3: Complete the square for the [tex]\(x\)[/tex]-terms:
For the quadratic expression [tex]\(x^2 + 2x\)[/tex], take half of the coefficient of [tex]\(x\)[/tex] (which is 2), square it, and add and subtract this square inside the equation.
[tex]\[ x^2 + 2x = (x + 1)^2 - 1 \][/tex]
Step 4: Complete the square for the [tex]\(y\)[/tex]-terms:
Similarly, for [tex]\(y^2 + 4y\)[/tex], take half of the coefficient of [tex]\(y\)[/tex] (which is 4), square it, and add and subtract this square inside the equation.
[tex]\[ y^2 + 4y = (y + 2)^2 - 4 \][/tex]
Step 5: Rewrite the equation with the completed squares:
[tex]\[ (x + 1)^2 - 1 + (y + 2)^2 - 4 = 20 \][/tex]
Add the constants (-1 and -4) to both sides of the equation to balance it:
[tex]\[ (x + 1)^2 + (y + 2)^2 - 5 = 20 \][/tex]
[tex]\[ (x + 1)^2 + (y + 2)^2 = 25 \][/tex]
This is now in standard form:
[tex]\[ (x + 1)^2 + (y + 2)^2 = 25 \][/tex]
### Part B: Identify the center and radius of the circle
Step 1: Identify the center [tex]\((h, k)\)[/tex]:
From the standard form of the circle equation [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex]:
Our equation is [tex]\((x + 1)^2 + (y + 2)^2 = 25\)[/tex].
Compare this with [tex]\( (x - (-1))^2 + (y - (-2))^2 = 25 \)[/tex]:
The center [tex]\((h, k)\)[/tex] is:
[tex]\[ (-1, -2) \][/tex]
Step 2: Identify the radius [tex]\(r\)[/tex]:
The right side of the equation is [tex]\(r^2 = 25\)[/tex]. The radius [tex]\(r\)[/tex] is the square root of 25:
[tex]\[ r = \sqrt{25} = 5 \][/tex]
Summary:
- The center of the circle is [tex]\((-1, -2)\)[/tex].
- The radius of the circle is [tex]\(5\)[/tex].