DeanBH said:
i have shown my working, i just tried to factorize in a way that would give me the co-efficients of x and y properly,
x^2 + y^2 + 2x -12y + 12 = 0
which is the same as x^2+ 2x+ y^2- 12y= -12
(x-a)^2 + (y-b)^2 = r^2 in that form.
why doesn't this work.
(x+1) + (y-6)
?? Did you mean (x+1)^2+ (y- 6)^2?
this gives you your x^2 your 2x and a +1 Gives you y^2 and -12y and a +36
i did (x + 1) because it gives you the 2x and the x^2 of the equation, but it also gives +1, i kept this on the left side of the equation so i ended up with +13 there. i then did the same for (y-6) and ended up with 36+13 on the left side of the equation which is 49.
Did you notice that you never do say
what equation you wound up with?
Okay, (x+ 1)^2+ (y- 6)^2= x^2+ 2x+ 1+ y^2- 12y+ 36= what?
Since, in order to "complete the square" with x^2+ 2x, you have to add 1, and, in order to "complete the square" with y^2- 12y, you have to add 36, you have to add add a total of 37.
You started with x^2+ 2x+ y^2- 12y= -12. Adding 37 to both sides of that gives
(x+ 1)^2+ (y- 6)^2= 37-12= 25. You have -12+ 1+ 36, not 12+ 36+ 1, because they are on the
right side of the equation.
The other way you have done this was, with your original equation, x^2+ y^2+ 2x- 12y+ 12= 0 is to recognize that you need a total of 36+ 1= 37 on left side. Since you already have 12, you need to add 37- 12= 25 to both sides:
x^2+ y^2+ 2x- 12y+ 12+ 25= 25 or (x^2+ 2x+ 1)+ (y^2- 12y+ 36)= (x+1)^2+ (y- 6)^2= 25.
