Is circle formula same in different XY system? I want to check an element if it's in a circle in my browser. I know the circle formula from my high school: 

But I'm not sure if it's same in browser XY system too. Because browsers zero coordinate is on left top of the page and from there to right is positive x axis and to bottom is positive y axis. 

For example in this picture point c have positive x and y.
Please optimize the circle formula for this system and also give me the correct formula for a circle centered in c = (200,200) with 150 radius. 
 A: For any affine change of variables $x \to a_1 x + a_2 y + a_3$ and $y \to b_1 x + b_2 y + b_3$ the equation of the circle $(x-x_0)^2 + (y-y_0)^2 = r^2$ translates into 
$$
   ( a_1 x + a_2 y + a_3 - x_0)^2 + (b_1 x + b_2 y +b_3 - y_0)^2 = r^2
$$
Expanding
$$
  \begin{eqnarray}
  & & x^2 ( a_1^2 + b_1^2) + y^2 (a_2^2 + b_2^2) + 2 x y ( a_1 a_2 + b_1 b_2) \\
 && +
  2 x ( a_1 (a_3 - x_0) + b_1 (b_3-y_0) ) + 
  2 y ( a_2 (a_3 - x_0) + b_2 (b_3-y_0) ) \\
 && + 
   (a_3-x_0)^2 + (b_3 - y_0)^2  = r^2 
  \end{eqnarray}
$$
Now, choosing $a_1$, $a_2$ and $b_1$ and $b_2$ so that 
$$
   a_1^2 + b_1^2 = a_2^2 + b_2^2 = 1  \qquad  a_1 a_2 + b_1 b_2 = 0
$$
and completing squares will again arrive at $(x-\hat{x}_0)^2 + (y-\hat{y}_0)^2 = \hat{r}^2$. 
A: Even with the $y$-coordinate flipped, the equation for your circle is still
$$
(x-200)^2+(y-200)^2=150^2\tag{1}
$$
In the normal coordinates, this would be the circle of radius $150$ and center $(200,-200)$ so the equation with flipped $y$-coordinate would be
$$
(x-200)^2+(-y-(-200))^2=150^2\tag{2}
$$
Equation $(2)$ is the same as equation $(1)$ since $(-1)^2=1$.
