To solve the problem of finding the center and radius of the given circle equation [tex]\( x^2 + y^2 + 14x + 2y + 14 = 0 \)[/tex], we need to rewrite it in the standard form of a circle equation [tex]\((x-h)^2 + (y-k)^2 = r^2\)[/tex].
Here are the steps:
1. Rewrite the equation:
Given: [tex]\( x^2 + y^2 + 14x + 2y + 14 = 0 \)[/tex]
2. Complete the square for the [tex]\(x\)[/tex] terms:
The [tex]\(x\)[/tex] terms are [tex]\( x^2 + 14x \)[/tex].
To complete the square, we add and subtract the square of half the coefficient of [tex]\(x\)[/tex] (which is 14/2 = 7):
[tex]\[
x^2 + 14x \rightarrow (x + 7)^2 - 49
\][/tex]
3. Complete the square for the [tex]\(y\)[/tex] terms:
The [tex]\(y\)[/tex] terms are [tex]\( y^2 + 2y \)[/tex].
To complete the square, we add and subtract the square of half the coefficient of [tex]\(y\)[/tex] (which is 2/2 = 1):
[tex]\[
y^2 + 2y \rightarrow (y + 1)^2 - 1
\][/tex]
4. Rewrite the given equation with these completions:
Substitute back into the original equation:
[tex]\[
(x + 7)^2 - 49 + (y + 1)^2 - 1 + 14 = 0
\][/tex]
5. Simplify the equation:
Combine the constant terms:
[tex]\[
(x + 7)^2 + (y + 1)^2 - 36 = 0
\][/tex]
Move the constant term to the other side to get the standard form:
[tex]\[
(x + 7)^2 + (y + 1)^2 = 36
\][/tex]
6. Identify the center and radius:
From the standard form [tex]\((x + 7)^2 + (y + 1)^2 = 36\)[/tex], we see that:
- The center [tex]\((h, k)\)[/tex] is at [tex]\((-7, -1)\)[/tex].
- The radius [tex]\(r\)[/tex] is [tex]\(\sqrt{36} = 6\)[/tex] units.
Based on these calculations, the correct answer is:
D. [tex]\((-7, -1), 6\)[/tex] units.
