Finding the intersection points of two circles - where one circle has a diff x and y coord than the other Following this link:
http://mathworld.wolfram.com/Circle-CircleIntersection.html
I now understand how to calculate the offset of the radical line from circle_a (a)
However:
Let two circles of radii  and  and centered at   and  intersect in a region shaped like an asymmetric lens. The equations of the two circles are

x^2 + y^2 = R^2
(x-d)^2 + y^2 = R^2

So this methods assumes that the y coordinates of both circles are the same?
How do I calculate the intersection points where both the x and y coordinates are not the same then?
Thanks.
 A: You can always assume that the y-coordinates of both circles are the same because you can rotate both circles in a way that they lie on a line that are parallel to the x-axis. Then you can translate both circles so the first one is in the origin.
Using that rotation the $d$ in your formula will be the distance of both centers and your formulas will look a lot better. To undo this you have to undo the translation and then the rotation. Either you use rotation matrices or you use a complete solution like for example this one.
A: to give an example, for instance $(x-1)^2+(y-1)^2=1, (x-2)^2+(y-2)^2=1$.  change variables, $(x,y)\mapsto(x+1,y+1)$ to get a new set of equations $x^2+y^2=1, (x-1)^2+(y-1)^2=1$.  the center of the first circle is now at the origin and the other is centered at $(1,1)$.  now the angle from the positive $x$-axis to $(1,1)$ is $\pi/4$ (in general you would have to find some inverse tangent; i picked an easy one).  so you want to rotate the whole plane by $-\pi/4$ ie
$$
(x,y)\mapsto
\left(
\begin{array}{cc}
1/\sqrt{2}&1/\sqrt{2}\\
-1/\sqrt{2}&1/\sqrt{2}\\
\end{array}
\right)
\left(
\begin{array}{c}
x\\
y\\
\end{array}
\right)
=\Bigg(\frac{(x+y)}{\sqrt{2}},\frac{(y-x)}{\sqrt{2}}\Bigg)
$$
now you have the equations $x^2+y^2=1, (x-\sqrt{2})^2+y^2=1$.
