To determine the center and radius of the circle given by the equation [tex]\((x-2)^2+(y-6)^2=64\)[/tex], we will follow these steps:
1. Identify the center of the circle:
- The general form of a circle's equation is [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.
- By comparing the given equation [tex]\((x-2)^2 + (y-6)^2 = 64\)[/tex] with the general form, we can identify the values of [tex]\(h\)[/tex] and [tex]\(k\)[/tex].
- Here, [tex]\(h = 2\)[/tex] and [tex]\(k = 6\)[/tex].
- Therefore, the center of the circle is [tex]\((2, 6)\)[/tex].
2. Determine the radius of the circle:
- In the general form [tex]\((x-h)^2 + (y-k)^2 = r^2\)[/tex], [tex]\(r^2\)[/tex] is the term on the right-hand side of the equation.
- For the given equation, [tex]\(r^2 = 64\)[/tex].
- To find the radius [tex]\(r\)[/tex], we take the square root of 64.
- Hence, [tex]\(r = \sqrt{64} = 8\)[/tex].
So, the center of the circle is [tex]\((2, 6)\)[/tex] and the radius is 8.
Among the provided options, the correct one is:
[tex]\[ \text{Center: } (2, 6); \text{ Radius: } 8 \][/tex]
