Please can you use English

The given equation is $x^2 + y^2 - 6x + 2y - 6 = 0$.

$$(x^2 - 6x) + (y^2 + 2y) = 6$$

To complete the square for $x^2 - 6x$, take half of the coefficient of $x$ (which is $-6/2 = -3$) and square it ($(-3)^2 = 9$).
To complete the square for $y^2 + 2y$, take half of the coefficient of $y$ (which is $2/2 = 1$) and square it ($1^2 = 1$).
Add these values to both sides of the equation:

$$(x^2 - 6x + 9) + (y^2 + 2y + 1) = 6 + 9 + 1$$ $$(x - 3)^2 + (y + 1)^2 = 16$$

The standard equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius.
Comparing our equation $(x - 3)^2 + (y + 1)^2 = 16$ with the standard form:
The center $(h, k)$ is $(3, -1)$.
The radius squared $r^2$ is 16, so the radius $r = \sqrt{16} = 4$.

The coordinates of the center are $(3, -1)$ and the radius is 4 units.

$$\text{Center: } (3, -1), \text{ Radius: } 4$$