To determine the coordinates of the center and the radius of the circle given by the equation [tex]\(x^2 + y^2 + 12x = -27\)[/tex], we need to rewrite the equation in the standard form [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex].
Here are the detailed steps:
1. Rewrite the equation and complete the square for the x terms:
The given equation is:
[tex]\[
x^2 + y^2 + 12x = -27
\][/tex]
2. Complete the square for the [tex]\(x\)[/tex] terms:
- Take the coefficient of [tex]\(x\)[/tex], which is 12, halve it to get 6, and then square it to get [tex]\(6^2 = 36\)[/tex].
- Add and subtract this square (36) inside the equation:
[tex]\[
x^2 + 12x + 36 + y^2 = -27 + 36
\][/tex]
3. Simplify and rewrite the equation:
- The left-hand side now becomes a perfect square trinomial, and the equation looks like:
[tex]\[
(x + 6)^2 + y^2 = 9
\][/tex]
Here, adding 36 on both sides keeps the equation balanced.
4. Identify the center and radius from the standard form:
- The standard form of a circle is:
[tex]\[
(x-h)^2 + (y-k)^2 = r^2
\][/tex]
Comparing this with [tex]\((x + 6)^2 + y^2 = 9\)[/tex], we see that:
- [tex]\(h = -6\)[/tex]
- [tex]\(k = 0\)[/tex]
- [tex]\(r^2 = 9\)[/tex]
Therefore, the center [tex]\((h, k)\)[/tex] is [tex]\((-6, 0)\)[/tex] and the radius [tex]\(r\)[/tex] is:
[tex]\[
r = \sqrt{9} = 3
\][/tex]
Thus the coordinates of the center of the circle are [tex]\((-6, 0)\)[/tex] and the radius is 3.
The correct answer is:
(3) center [tex]\((-6, 0)\)[/tex] and radius 3
