To find the radius of the circle given by the equation [tex]\(x^2 + y^2 - 2x + 8y - 47 = 0\)[/tex], we need to rewrite the equation 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.
### Step-by-Step Solution:
1. Group [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms:
We start with the given equation:
[tex]\[
x^2 + y^2 - 2x + 8y - 47 = 0
\][/tex]
2. Complete the square for the [tex]\(x\)[/tex] terms:
The [tex]\(x\)[/tex] terms are [tex]\(x^2 - 2x\)[/tex].
[tex]\[
x^2 - 2x = (x - 1)^2 - 1
\][/tex]
3. Complete the square for the [tex]\(y\)[/tex] terms:
The [tex]\(y\)[/tex] terms are [tex]\(y^2 + 8y\)[/tex].
[tex]\[
y^2 + 8y = (y + 4)^2 - 16
\][/tex]
4. Substitute the completed squares back into the original equation:
[tex]\[
(x - 1)^2 - 1 + (y + 4)^2 - 16 - 47 = 0
\][/tex]
Simplify the constants:
[tex]\[
(x - 1)^2 + (y + 4)^2 - 64 = 0
\][/tex]
[tex]\[
(x - 1)^2 + (y + 4)^2 = 64
\][/tex]
5. Identify the standard form:
Now, the equation is in the form [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex], where:
[tex]\[
(x - 1)^2 + (y + 4)^2 = 64
\][/tex]
Here, [tex]\(h = 1\)[/tex], [tex]\(k = -4\)[/tex], and [tex]\(r^2 = 64\)[/tex].
6. Find the radius [tex]\(r\)[/tex]:
[tex]\[
r = \sqrt{64} = 8
\][/tex]
### Conclusion:
The radius of the circle is:
[tex]\[
\boxed{8} \, \text{units}
\][/tex]
