Define "y" value from the equation of circle Let's take a circle. It has the following general equation to describe it:
$(x-u)^2+(y-v)^2=r^2$
,where $u,v$ is the coordinates of the center of the circle, and $r$ is the radius of the circle. If I could understand it well: all those points with coordinates $x,y$ which makes this equation true are lying on this circle exactly.
First quetion: am I right?
Also, I want to define $y$ from this equation, and I got this:
$y^2=r^2-x^2+2xu-u^2-2yv-v^2$
Second question: is this formule right? I'm not sure that I didn't screw this up. If it's wrong, what's the right form of defining $y$ from the general formula?
 A: You are right on the first question.
For the second, you have not really isolated $y$. For note that you have $y$'s on both sides of your equation.You could write $(y-v)^2=r^2-(x-u)^2$ and then
$$y=v\pm\sqrt{r^2-(x-u)^2}.$$
The solution with the $+$ gives the equation of the top half of the circle, and the solution with the $-$ gives the equation of the bottom half.
A: Yes you are correct, all the points $x,y$ which make this equation true are a distance $r$ away from the center, or on the circle. Second it should be $+2yv$ but you want to isolate $y$ so leave $y^2-2yv$ on the left hand side then you have $y^2-2yv = r^2-x^2+2xu-u^2-v^2$. Call the right hand side some constant $c$ and then use the quadratic equation to simplify. 
A: You are correct. In coordinate geometry, we have equations, and if a point's coordinate makes the equation true, it is part of the graph. The equation's graph is a circle. This is because, first of all, the $x$ distance from the point $(u, v$) is $|x-u|$ and the $y$ distance is $|y-v|$. $r$ is the total distance. by Pythagorean theorem you get.
$$|x-u|^2+|y-v|^2=r^2$$
$$(x-u)^2+(y-v)^2=r^2$$
Also, there is a slight sign error in your second equation. It is actually:
$$y^2=r^2-x^2+2xu-u^2+2yv-v^2$$
Note though this does not define $y$. This will tell us what $y^2$ is, not $y$. Also, it uses $y$ in its definition. To get $y$, we do:
$$y^2-2yv=r^2-x^2+2xu-u^2-v^2$$
$$y^2-2yv+v^2=r^2-x^2+2xu-u^2-v^2+v^2$$
$$(y-v)^2=r^2-x^2+2xu-u^2$$
$$y-v=\pm\sqrt{r^2-x^2+2xu-u^2}$$
$$y=v\pm\sqrt{r^2-x^2+2xu-u^2}$$
This will tell us what $y$ is. There are actually two possible $y$ values for each $x$. You can see this if you look at a circle. It should also be noted that even though this is correct equation for $y$, people usually write it in a simpler way. You can see this from the other answers.
