To identify the center and the radius of the circle given by the equation [tex]\((x-5)^2 + y^2 = 81\)[/tex], we can compare it with the standard form of the equation of a circle, which is:
[tex]\[
(x-h)^2 + (y-k)^2 = r^2
\][/tex]
Here, [tex]\((h, k)\)[/tex] represents the center of the circle, and [tex]\(r\)[/tex] is the radius.
1. Identify the center of the circle:
The given equation is:
[tex]\[
(x-5)^2 + y^2 = 81
\][/tex]
On comparing this with the standard form [tex]\((x-h)^2 + (y-k)^2 = r^2\)[/tex],
- The value of [tex]\(h\)[/tex] is 5, as indicated by the term [tex]\((x-5)\)[/tex].
- The value of [tex]\(k\)[/tex] is 0, as indicated by the term [tex]\(y^2\)[/tex] (which can be seen as [tex]\((y-0)^2\)[/tex]).
Therefore, the center of the circle is [tex]\((h, k) = (5, 0)\)[/tex].
2. Identify the radius of the circle:
In the standard form, [tex]\(r^2\)[/tex] is the constant term on the right side of the equation.
For the given equation:
[tex]\[
(x-5)^2 + y^2 = 81
\][/tex]
- The right side of the equation is 81.
- We can compare this with [tex]\(r^2\)[/tex] to find the radius [tex]\(r\)[/tex].
Therefore,
[tex]\[
r^2 = 81
\][/tex]
Taking the square root on both sides,
[tex]\[
r = \sqrt{81} = 9
\][/tex]
Thus, the radius of the circle is 9 units.
To summarize:
- The center of the circle is at [tex]\((5, 0)\)[/tex].
- The radius of the circle is 9 units.
So, we fill in the blanks as follows:
The radius of the circle is [tex]\(9\)[/tex] units.
The center of the circle is at [tex]\((5, 0)\)[/tex].
