# 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

Update:

i know where i went wrong, thank you!

Relevance
• TC
Lv 7
10 years ago

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

• 10 years ago

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:

xÂ² + yÂ² - 2x + 6y + 1 = 0

xÂ² - 2x + yÂ² + 6y = -1

(x - 1)Â² - 1 + (y + 3)Â² - 9 = -1

(x - 1)Â² + (y + 3)Â² = 9