3.16. Reduce the equation of the circle $x^{2}-2 x+y^{2}+$ $+6 y=6$ to its canonical form. Determine the coordinates of the center and the radius.
Solution. We use the formulas for the complete square of the sum and difference, and complete the square of the expressions

a) $x^{2}-2 x=x^{2}-2 \cdot x \cdot 1+1^{2}-1^{2}=(x-1)^{2}-1$;

b) $y^{2}+6 y=y^{2}-2 \cdot y \cdot 3+3^{2}-3^{2}=(y+3)^{2}-9$.

The equation of the circle can now be rewritten as follows:

$$
(x-1)^{2}-1+(y+3)^{2}-9=6 \text {, or }(x-1)^{2}+(y+3)^{2}=4^{2} \text {. }
$$

The center of the circle $O(1 ;-3)$, radius $R=4$.