Hi ADY,

Your solution (however wrong) seems to be a bit convoluted.

Look at it simply.

$\displaystyle x^2-10x+y^2+8y+36=0$

First, group your x and y, and move +36 to the right side.

$\displaystyle x^2-10x+y^2+8y=-36$

The trick here is to add a constant to the x and y group that will make the resulting trinomial a perfect square.

How do we do that?

Take half of the coefficient of the linear term, square it, then add it to the group. Do that for both the x and y group. And since we added a certain amount to the left side, we must add that same amount to the right side.

So, here we go.

Half of -10 is -5. We square that to get 25 and we add that to the x group to get $\displaystyle x^2-10x{\color{blue}+25}$.

Half of 8 is 4. We square that to get 16 and we add that to the y group to get $\displaystyle y^2+8x{\color{red}+16}$.

We added a total of 25 + 16 on the left, so we add that amount to the right to keep things balanced.

$\displaystyle (x^2-10x {\color{blue}+25})+(y^2+8x{\color{red}+16})=-36{\color{blue}+25}{\color{red}+16}$

Finally, grouping the two perfect square trinomials, we arrive at

$\displaystyle \boxed{(x-5)^2+(y+4)^2=5}$

Center (5, -4), Radius = $\displaystyle \sqrt{5}$