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Sagot :
To rewrite the equation [tex]\( x^2 + 6x + y^2 + 8y + 24 = 0 \)[/tex] in the center-radius form of the equation of a circle, we need to complete the square for the [tex]\(x\)[/tex]-terms and the [tex]\(y\)[/tex]-terms. Let's go through the steps in detail:
1. Rewrite the Equation:
[tex]\( x^2 + 6x + y^2 + 8y + 24 = 0 \)[/tex]
2. Group the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] Terms Separately:
[tex]\( (x^2 + 6x) + (y^2 + 8y) + 24 = 0 \)[/tex]
3. Complete the Square for the [tex]\(x\)[/tex]-Terms:
To complete the square for [tex]\(x^2 + 6x\)[/tex]:
- Take half of the coefficient of [tex]\(x\)[/tex], which is 3, and square it: [tex]\( (6/2)^2 = 3^2 = 9 \)[/tex].
- Add and subtract 9 inside the parentheses:
[tex]\( (x^2 + 6x + 9 - 9) = (x + 3)^2 - 9 \)[/tex]
4. Complete the Square for the [tex]\(y\)[/tex]-Terms:
To complete the square for [tex]\(y^2 + 8y\)[/tex]:
- Take half of the coefficient of [tex]\(y\)[/tex], which is 4, and square it: [tex]\( (8/2)^2 = 4^2 = 16 \)[/tex].
- Add and subtract 16 inside the parentheses:
[tex]\( (y^2 + 8y + 16 - 16) = (y + 4)^2 - 16 \)[/tex]
5. Substitute the Completed Squares Back into the Equation:
Replace the original expressions with the completed squares:
[tex]\( (x + 3)^2 - 9 + (y + 4)^2 - 16 + 24 = 0 \)[/tex]
6. Combine and Simplify Constant Terms:
Combine [tex]\(-9\)[/tex], [tex]\(-16\)[/tex], and [tex]\(24\)[/tex]:
[tex]\( -9 - 16 + 24 = -1 \)[/tex]
Thus, the equation becomes:
[tex]\( (x + 3)^2 + (y + 4)^2 - 1 = 0 \)[/tex]
7. Rearrange the Equation:
Add 1 to both sides to isolate the completed squares:
[tex]\( (x + 3)^2 + (y + 4)^2 = 1 \)[/tex]
This is now in the center-radius form of a circle, [tex]\((x-h)^2 + (y-k)^2 = r^2\)[/tex], with center [tex]\((-3, -4)\)[/tex] and radius [tex]\(1\)[/tex].
Comparing this with the given options, we see that the correct answer is:
D. [tex]\((x + 3)^2 + (y + 4)^2 = 1\)[/tex]
1. Rewrite the Equation:
[tex]\( x^2 + 6x + y^2 + 8y + 24 = 0 \)[/tex]
2. Group the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] Terms Separately:
[tex]\( (x^2 + 6x) + (y^2 + 8y) + 24 = 0 \)[/tex]
3. Complete the Square for the [tex]\(x\)[/tex]-Terms:
To complete the square for [tex]\(x^2 + 6x\)[/tex]:
- Take half of the coefficient of [tex]\(x\)[/tex], which is 3, and square it: [tex]\( (6/2)^2 = 3^2 = 9 \)[/tex].
- Add and subtract 9 inside the parentheses:
[tex]\( (x^2 + 6x + 9 - 9) = (x + 3)^2 - 9 \)[/tex]
4. Complete the Square for the [tex]\(y\)[/tex]-Terms:
To complete the square for [tex]\(y^2 + 8y\)[/tex]:
- Take half of the coefficient of [tex]\(y\)[/tex], which is 4, and square it: [tex]\( (8/2)^2 = 4^2 = 16 \)[/tex].
- Add and subtract 16 inside the parentheses:
[tex]\( (y^2 + 8y + 16 - 16) = (y + 4)^2 - 16 \)[/tex]
5. Substitute the Completed Squares Back into the Equation:
Replace the original expressions with the completed squares:
[tex]\( (x + 3)^2 - 9 + (y + 4)^2 - 16 + 24 = 0 \)[/tex]
6. Combine and Simplify Constant Terms:
Combine [tex]\(-9\)[/tex], [tex]\(-16\)[/tex], and [tex]\(24\)[/tex]:
[tex]\( -9 - 16 + 24 = -1 \)[/tex]
Thus, the equation becomes:
[tex]\( (x + 3)^2 + (y + 4)^2 - 1 = 0 \)[/tex]
7. Rearrange the Equation:
Add 1 to both sides to isolate the completed squares:
[tex]\( (x + 3)^2 + (y + 4)^2 = 1 \)[/tex]
This is now in the center-radius form of a circle, [tex]\((x-h)^2 + (y-k)^2 = r^2\)[/tex], with center [tex]\((-3, -4)\)[/tex] and radius [tex]\(1\)[/tex].
Comparing this with the given options, we see that the correct answer is:
D. [tex]\((x + 3)^2 + (y + 4)^2 = 1\)[/tex]
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