What are the center and radius of the circle defined by the equation [tex]x^2 + y^2 - 6x + 10y + 25 = 0[/tex]?

A. Center [tex]\((3, -5)\)[/tex]; radius 3
B. Center [tex]\((-3, 5)\)[/tex]; radius 3
C. Center [tex]\((3, -5)\)[/tex]; radius 9
D. Center [tex]\((-3, 5)\)[/tex]; radius 9



Answer :

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]

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