To determine the center and the radius of the circle from the given equation [tex]\(x^2 + y^2 + 14x + 2y + 14 = 0\)[/tex], we need to rewrite the equation into the standard form of a circle.
The standard form of a circle's equation is:
[tex]\[
(x - h)^2 + (y - k)^2 = r^2
\][/tex]
where [tex]\((h, k)\)[/tex] is the center of the circle and [tex]\(r\)[/tex] is the radius.
Let's start by completing the square for both [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms in the given equation:
1. Group the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms:
[tex]\[
x^2 + 14x + y^2 + 2y + 14 = 0
\][/tex]
2. Complete the square for the [tex]\(x\)[/tex] terms:
[tex]\[
x^2 + 14x \quad \Rightarrow \quad (x + 7)^2 - 49
\][/tex]
Here, we add and subtract 49 to complete the square.
3. Complete the square for the [tex]\(y\)[/tex] terms:
[tex]\[
y^2 + 2y \quad \Rightarrow \quad (y + 1)^2 - 1
\][/tex]
Here, we add and subtract 1 to complete the square.
4. Rewrite the equation with the completed squares:
[tex]\[
(x + 7)^2 - 49 + (y + 1)^2 - 1 + 14 = 0
\][/tex]
Simplify this form:
[tex]\[
(x + 7)^2 + (y + 1)^2 - 49 - 1 + 14 = 0 \quad \Rightarrow \quad (x + 7)^2 + (y + 1)^2 - 36 = 0
\][/tex]
[tex]\[
(x + 7)^2 + (y + 1)^2 = 36
\][/tex]
Now the equation is in the standard form [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex], where [tex]\((h, k)\)[/tex] is the center and [tex]\(r\)[/tex] is the radius.
- The center [tex]\((h, k)\)[/tex] is [tex]\((-7, -1)\)[/tex].
- The radius [tex]\(r\)[/tex] is [tex]\(\sqrt{36} = 6\)[/tex] units.
Thus, the correct answer is:
D. [tex]\((-7, -1), 6\)[/tex] units.
