To determine which graph represents the equation [tex]\((x-3)^2 + (y+1)^2 = 9\)[/tex], we need to recognize important characteristics of this equation. This equation is in the standard form of a circle:
[tex]\[
(x-h)^2 + (y-k)^2 = r^2
\][/tex]
where [tex]\((h, k)\)[/tex] is the center of the circle and [tex]\(r\)[/tex] is the radius.
1. Identify the Center:
The terms [tex]\((x-3)^2\)[/tex] and [tex]\((y+1)^2\)[/tex] indicate a translation of the circle's center from the origin [tex]\((0,0)\)[/tex] to [tex]\((3, -1)\)[/tex]. This is because the general form of the circle [tex]\((x-h)^2\)[/tex] implies the center is shifted [tex]\(h\)[/tex] units right if [tex]\(h\)[/tex] is positive, and [tex]\((y-k)^2\)[/tex] implies the center is shifted [tex]\(k\)[/tex] units up if [tex]\(k\)[/tex] is positive. Here, [tex]\(h\)[/tex] is 3 and [tex]\(k\)[/tex] is -1.
Center: [tex]\((3, -1)\)[/tex]
2. Calculate the Radius:
The right-hand side of the equation [tex]\(9\)[/tex] is represented as [tex]\(r^2\)[/tex]. To find the radius [tex]\(r\)[/tex], we take the square root of [tex]\(9\)[/tex]:
[tex]\[
r = \sqrt{9} = 3
\][/tex]
Radius: [tex]\(3\)[/tex]
3. Conclusion:
The graph of the given equation [tex]\((x-3)^2 + (y+1)^2 = 9\)[/tex] is a circle centered at [tex]\((3, -1)\)[/tex] with a radius of [tex]\(3\)[/tex].
When identifying the correct graph among available options, look for a circle with:
- A center at the point [tex]\((3, -1)\)[/tex]
- A radius extending 3 units in all directions from the center.
Thus, the graph will depict a circle positioned such that its center is at [tex]\((3, -1)\)[/tex] and the circle expands 3 units in every direction from this center point.
