To find the radius of the circle given by the equation \( x^2 + y^2 + 8x - 6y + 21 = 0 \), we'll follow these steps:
1. Rewrite the equation in the standard form of a circle's equation by completing the square.
2. Group the \( x \) terms and \( y \) terms together:
[tex]\[
(x^2 + 8x) + (y^2 - 6y) = -21
\][/tex]
3. Complete the square for the \( x \) terms and \( y \) terms separately:
- For the \( x \) terms, \( x^2 + 8x \):
[tex]\[
x^2 + 8x = (x + 4)^2 - 16
\][/tex]
- For the \( y \) terms, \( y^2 - 6y \):
[tex]\[
y^2 - 6y = (y - 3)^2 - 9
\][/tex]
4. Substitute these completed squares back into the equation:
[tex]\[
(x + 4)^2 - 16 + (y - 3)^2 - 9 = -21
\][/tex]
5. Simplify the equation:
[tex]\[
(x + 4)^2 + (y - 3)^2 - 25 = -21
\][/tex]
6. Move the constant term to the right side of the equation:
[tex]\[
(x + 4)^2 + (y - 3)^2 = 4
\][/tex]
7. Recognize this as the standard form of a circle's equation \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center of the circle and \( r \) is the radius.
Here, \((x + 4)^2 + (y - 3)^2 = 4\), so:
[tex]\[
\text{Radius squared} = 4
\][/tex]
[tex]\[
\text{Radius} = \sqrt{4} = 2
\][/tex]
Thus, the radius of the circle is 2 units. The correct option is:
```plaintext
2 units
```
