To solve for the center and the radius of the circle given by the equation [tex]\( x^2 + y^2 - 6x + 8y = 0 \)[/tex], let's rewrite it in standard form [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex].
1. Completing the Square:
- For the [tex]\( x \)[/tex]-terms [tex]\( x^2 - 6x \)[/tex]:
[tex]\[
x^2 - 6x \quad \text{can be rewritten by completing the square as}\quad (x - 3)^2 - 9
\][/tex]
- For the [tex]\( y \)[/tex]-terms [tex]\( y^2 + 8y \)[/tex]:
[tex]\[
y^2 + 8y \quad \text{can be rewritten by completing the square as}\quad (y + 4)^2 - 16
\][/tex]
2. Rewrite the equation:
[tex]\[
x^2 - 6x + y^2 + 8y = 0
\][/tex]
Substituting the completed squares:
[tex]\[
(x - 3)^2 - 9 + (y + 4)^2 - 16 = 0
\][/tex]
3. Simplify:
[tex]\[
(x - 3)^2 + (y + 4)^2 - 25 = 0
\][/tex]
Adding 25 to both sides:
[tex]\[
(x - 3)^2 + (y + 4)^2 = 25
\][/tex]
Now we have the circle in standard form:
[tex]\[
(x - h)^2 + (y - k)^2 = r^2
\][/tex]
By comparing, we can identify:
- [tex]\( h = 3 \)[/tex]
- [tex]\( k = -4 \)[/tex]
- [tex]\( r^2 = 25 \)[/tex] so [tex]\( r = \sqrt{25} = 5 \)[/tex]
Thus, the center of the circle is [tex]\((3, -4)\)[/tex] and the radius is [tex]\(5\)[/tex] units.
The correct answer is:
D. [tex]\((3, -4), 5\)[/tex] units
