To find the center and radius of the circle defined by the equation [tex]\( x^2 + y^2 - 6x + 10y + 25 = 0 \)[/tex], we need to rewrite the equation in the standard form of a circle's 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.
Here are the detailed steps:
1. Rewrite the equation by grouping the [tex]\(x\)[/tex]-terms and [tex]\(y\)[/tex]-terms:
[tex]\[
x^2 - 6x + y^2 + 10y + 25 = 0
\][/tex]
2. Complete the square for the [tex]\(x\)[/tex]-terms:
[tex]\[
x^2 - 6x \quad \Rightarrow \quad x^2 - 6x + 9 - 9 \quad \Rightarrow \quad (x - 3)^2 - 9
\][/tex]
3. Complete the square for the [tex]\(y\)[/tex]-terms:
[tex]\[
y^2 + 10y \quad \Rightarrow \quad y^2 + 10y + 25 - 25 \quad \Rightarrow \quad (y + 5)^2 - 25
\][/tex]
4. Substitute back into the equation:
[tex]\[
(x - 3)^2 - 9 + (y + 5)^2 - 25 + 25 = 0
\][/tex]
5. Simplify the equation:
[tex]\[
(x - 3)^2 - 9 + (y + 5)^2 = 0
\][/tex]
[tex]\[
(x - 3)^2 + (y + 5)^2 - 9 = 0
\][/tex]
[tex]\[
(x - 3)^2 + (y + 5)^2 = 9
\][/tex]
6. Rewrite in the standard circle form [tex]\( (x - h)^2 + (y - k)^2 = r^2 \)[/tex]:
[tex]\[
(x - 3)^2 + (y + 5)^2 = 3^2
\][/tex]
From this, we can see that the center of the circle [tex]\((h, k)\)[/tex] is [tex]\((3, -5)\)[/tex] and the radius [tex]\(r\)[/tex] is 3.
Thus, the correct answer is:
[tex]\[ \boxed{A \text{. Center } (3, -5) ; \text{ radius } 3} \][/tex]
