The equation of a circle is [tex][tex]$x^2 + y^2 + 12x = -27$[/tex][/tex]. What are the coordinates of the center and the length of the radius of the circle?

1. Center [tex][tex]$(6, 0)$[/tex][/tex] and radius 3
2. Center [tex][tex]$(6, 0)$[/tex][/tex] and radius 9
3. Center [tex][tex]$(-6, 0)$[/tex][/tex] and radius 3
4. Center [tex][tex]$(-6, 0)$[/tex][/tex] and radius 9



Answer :

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