Pi for non mathematician I've been long gone from math (shamefully) and have trouble using some quite familiar concepts...
Consider the following picture in which I render two circles with radius 32 (2) on a plane of 1024 by 1024 (1). 
I chose a point at which I render the first circle [x,y] (3). 
I choose to render another circle exactly 1 radius above the first circle.
How do I calculate 4? Which is the intersection of the two circles? (can be left or right side). 

I understand I can calculate the circumference and divide it by six to get the distance from the center of the upper circle to the intersection. But I need coordinates. 

Any guide will be greatly appreciated!

 A: Call the center of the first circle $(x_1, y_1)$. And call the radius $r$ (which is 32, but my answer will be more general). The center of the second circle is $(x_2, y_2)$. You want to know the location of the intersections between the two circles. Here goes:


*

*$y_2 = y_1 + r$, because you've put the second circle one radius higher. And $x_1 = x_2$, because the two centers are vertically aligned. 

*The intersection point must be at height $y_3 = y_1 + \frac{r}{2}$, because it's halfway (in height) between $y_1$ and $y_2$. 

*The intersection point $(x_3, y_3)$ must have distance $r$ to the two circle centers. Writing that for the first circle center, we have
$$
\sqrt{ (x_3 - x_1)^2  + (y_3 - y_1)^2 } = r = \sqrt{r^2}.
$$
which comes, from a long chain of steps, from Pythagoras' Theorem.
If two numbers have the same square root, they're the same number, so 
$$
(x_3 - x_1)^2  + (y_3 - y_1)^2  = r^2
$$
Now by step 2, we know that $y_3 - y_1$ is $r/2$,  So we can say
$$
 (a - x_1)^2  + (r/2)^2  = r^2\\
 (a - x_1)^2  + r^2/4  = r^2 \\
 (a - x_1)^2  = \frac{3}{4} r^2.
$$
That means that $(a - x_1)$ must be either $\sqrt{\frac{3}{4}} r$ or its negative. 
In your case, this means that the  intersection point is 
$$
\sqrt{ \frac{3}{4} } 32 \approx 27.71
$$
to the left or right  of your circle centers, and the $y$-coordinate is 16 units greater than the first circle center. So 
$$
(x_3, y_3) \approx (x_1 \pm 27.71, y_1 + 16). 
$$ 
