To match the standard form of an equation for a circle with the given centers and radii, we need to pair each equation to its respective center and radius. Here’s how:
1. Center: [tex]\((6, 3)\)[/tex], Radius: [tex]\(2\)[/tex]
- The standard form of the equation for a circle is [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.
- Substituting [tex]\((h, k) = (6, 3)\)[/tex] and [tex]\(r = 2\)[/tex], we get [tex]\((x-6)^2 + (y-3)^2 = 4\)[/tex].
Thus, the equation is [tex]\((x-6)^2 + (y-3)^2 = 4\)[/tex].
2. Center: [tex]\((6, -3)\)[/tex], Radius: [tex]\(2\)[/tex]
- Using the same standard form, substituting [tex]\((h, k) = (6, -3)\)[/tex] and [tex]\(r = 2\)[/tex] gives [tex]\((x-6)^2 + (y+3)^2 = 4\)[/tex].
Thus, the equation is [tex]\((x-6)^2 + (y+3)^2 = 4\)[/tex].
3. Center: [tex]\((-3, 6)\)[/tex], Radius: [tex]\(4\)[/tex]
- Substituting [tex]\((h, k) = (-3, 6)\)[/tex] and [tex]\(r = 4\)[/tex] into the standard form of the circle equation gives [tex]\((x+3)^2 + (y-6)^2 = 16\)[/tex].
Thus, the equation is [tex]\((x+3)^2 + (y-6)^2 = 16\)[/tex].
4. Center: [tex]\((3, -6)\)[/tex], Radius: [tex]\(4\)[/tex]
- Substituting [tex]\((h, k) = (3, -6)\)[/tex] and [tex]\(r = 4\)[/tex], we get [tex]\((x-3)^2 + (y+6)^2 = 16\)[/tex].
Thus, the equation is [tex]\((x-3)^2 + (y+6)^2 = 16\)[/tex].
Therefore, the correct pairs are:
- Center: [tex]\((6, 3)\)[/tex], Radius: [tex]\(2\)[/tex] ⟷ [tex]\((x-6)^2 + (y-3)^2 = 4\)[/tex]
- Center: [tex]\((6, -3)\)[/tex], Radius: [tex]\(2\)[/tex] ⟷ [tex]\((x-6)^2 + (y+3)^2 = 4\)[/tex]
- Center: [tex]\((-3, 6)\)[/tex], Radius: [tex]\(4\)[/tex] ⟷ [tex]\((x+3)^2 + (y-6)^2 = 16\)[/tex]
- Center: [tex]\((3, -6)\)[/tex], Radius: [tex]\(4\)[/tex] ⟷ [tex]\((x-3)^2 + (y+6)^2 = 16\)[/tex]
