1. ## circle equation

i'm looking at a worked example and it says:

a circle has equation $\displaystyle x^2 + y^2 - 8x + 10y = 0$ it says the solution involves writing the equation of the circle as $\displaystyle [x^2 - 8x] + [y^2 + 10y] = 0$ which i understand. it then goes on to talk about completing the square and makes it $\displaystyle [x^2 - 8x + (-4)^2] + [y^2 + 10y + (5)^2] = (-4)^2 + (5)^2$ and then
$\displaystyle (x - 4)^2 + (y + 5)^2 = 16 + 25$

i don't see why the $\displaystyle (-4)^2$ and the $\displaystyle (5)^2$ don't become $\displaystyle -(-4)^2$ and $\displaystyle -(5)^2$ when they've been moved to the other side of the = sign?

any help would be appreciated

2. Originally Posted by mark
i'm looking at a worked example and it says:

a circle has equation $\displaystyle x^2 + y^2 - 8x + 10y = 0$ it says the solution involves writing the equation of the circle as $\displaystyle [x^2 - 8x] + [y^2 + 10y] = 0$ which i understand. it then goes on to talk about completing the square and makes it $\displaystyle [x^2 - 8x + (-4)^2] + [y^2 + 10y + (5)^2] = (-4)^2 + (5)^2$ and then
$\displaystyle (x - 4)^2 + (y + 5)^2 = 16 + 25$

i don't see why the $\displaystyle (-4)^2$ and the $\displaystyle (5)^2$ don't become $\displaystyle -(-4)^2$ and $\displaystyle -(5)^2$ when they've been moved to the other side of the = sign?

any help would be appreciated
Hi mark,

You are actually adding 16 and 25 to the left side of the equal sign in order to complete the square. So, you must add the same amount to the right side to maintain balance. That's why you wouldn't use $\displaystyle -(-4)^2 \ \ and \ \ -(5)^2$. If you did, you would be subtracting them from the right side.