To find the center and radius of the circle described by the equation [tex]\(x^2 + y^2 - 6x + 8y = 0\)[/tex], we need to rewrite the given equation in the standard form of a circle equation [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.
Let's go through the steps:
1. Rewrite the equation by completing the square for both [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms.
- For the [tex]\(x\)[/tex] terms [tex]\(x^2 - 6x\)[/tex]:
[tex]\[
x^2 - 6x \text{ can be written as } (x - 3)^2 - 9
\][/tex]
- For the [tex]\(y\)[/tex] terms [tex]\(y^2 + 8y\)[/tex]:
[tex]\[
y^2 + 8y \text{ can be written as } (y + 4)^2 - 16
\][/tex]
2. Substitute these completed squares back into the equation:
[tex]\[
x^2 + y^2 - 6x + 8y = (x - 3)^2 - 9 + (y + 4)^2 - 16 = 0
\][/tex]
3. Simplify the equation by moving the constant terms to the right side:
[tex]\[
(x - 3)^2 - 9 + (y + 4)^2 - 16 = 0
\][/tex]
[tex]\[
(x - 3)^2 + (y + 4)^2 - 25 = 0
\][/tex]
[tex]\[
(x - 3)^2 + (y + 4)^2 = 25
\][/tex]
4. Identify the components of the standard form equation:
- The term [tex]\((x - 3)^2\)[/tex] indicates that [tex]\(h = 3\)[/tex].
- The term [tex]\((y + 4)^2\)[/tex] indicates that [tex]\(k = -4\)[/tex].
- The right side of the equation [tex]\(25\)[/tex] represents [tex]\(r^2\)[/tex], giving [tex]\(r = \sqrt{25} = 5\)[/tex] units.
So, the center of the circle is [tex]\((3, -4)\)[/tex] and the radius is [tex]\(5\)[/tex] units. Therefore, the correct answer is:
B. [tex]\((3, -4), 5\)[/tex] units
