We complete the square for both $x$ and $y$.
For $x^2 - 6x$, we add $(6/2)^2 = 9$ to both sides.
For $y^2 + 2y$, we add $(2/2)^2 = 1$ to both sides.
This gives us $(x^2 - 6x + 9) + (y^2 + 2y + 1) = -6 + 9 + 1$,
or $(x - 3)^2 + (y + 1)^2 = 4$.
This is the equation of a circle with center $(3, -1)$ and radius $\sqrt{4} = \boxed{2}$.
The answer is: 2
Below is a math problem. Provide a detailed, step-by-step solution.

### Problem:
Find the radius of the circle with equation $x^2 - 6x + y^2 + 2y + 6 = 0$.

### Answer:
