Answer :

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
```

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