Can someone check this for me?--completing the square

its completing the square for both x & y for a circle ( (x-h)^2+(y-k)^2=r^2 )
this is the equation: x^2+y^2-2x+6y+1=0
the answer i got was (x+1)^2 (y+3) =9
my peer got (x-1)^2 (y+3) =9

i just want to know what im doing wrong, thanks SO much for the help

x^2 + y^2 - 2x + 6y + 1 = 0
(x^2 - 2x) + (y^2 + 6y) = - 1

complete sqaure of x

(- 2/2)^2 = (- 1)^2 = 1

complete square of y

(6/2)^2 = (3)^2 = 9

add (9 + 1) to both sides of equation

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

center is (1, - 3) with radius of 3

There is no need to add a constant before completing the square as the above answerers have done, you can complete the square straight away.

Complete the square for this circle equation:
x² + y² - 2x + 6y + 1 = 0
x² - 2x + y² + 6y = -1
(x - 1)² - 1 + (y + 3)² - 9 = -1
(x - 1)² + (y + 3)² = 9