How can I find the points at which two circles intersect? Given the radius and $x,y$ coordinates of the center point of two circles how can I calculate their points of intersection if they have any?
This can be done without any trigonometry at all. Let the equations of the circles be
$$(x-x_1)^2 + (y-y_1)^2 = r_1^2, \tag{1}$$
$$(x-x_2)^2 + (y-y_2)^2 = r_2^2. \tag{2}$$
By subtracting $(2)$ from $(1)$ and then expanding, we in fact obtain a linear equation for $x$ and $y$; after a little rearranging it becomes
$$-2x(x_1 - x_2) - 2y(y_1 - y_2) = (r_1^2 - r_2^2) - (x_1^2 - x_2^2) - (y_1^2 - y_2^2).$$
(If the circles intersect, this is the equation of the line that passes through the intersection points.) This equation can be solved for one of $x$ or $y$; let's suppose $y_1 - y_2 \ne 0$ so that we can solve for $y$:
$$y = -\frac{x_1 - x_2}{y_1 - y_2} x + \dotsc. \tag{3}$$
 Substituting this expression for $y$ into $(1)$ or $(2)$ gives a quadratic equation in only $x$. Then the $x$-coordinates of the intersection points are the solutions to this; the $y$-coordinates can be obtained by plugging the $x$-coordinates into $(3)$.
following the very nice hint of arkeet you can after defining a,b,c,g,h  using rc1,zc1,rc2,zc2, as the center point of the circles and R1 and R2 their radii
and assuming that rc2!=rc1 (the case where rc2=rc1 can be also done in the manner or arkeet, however the present assumption was useful to me)

g = (zc2 - zc1)/(rc2 - rc1);
h = (R1*R1-R2*R2 + zc2*zc2 - zc1*zc1 + rc2*rc2 - rc1*rc1 )/(2*(rc2 - rc1));

a = (g*g+ 1);
b = -2*g*h + 2*g*rc1 - 2*zc1;
c = h*h -2*h*rc1 + rc1*rc1 + zc1*zc1 - R1*R1;

zplus  = (-b +sqrt(b*b-4*a*c)) / (2*a); 
zminus = (-b -sqrt(b*b-4*a*c)) / (2*a); 
and
rplus  = -gzplus+h;
rminus = -gzminus+h;

you can test this by verifying that the intersection points do lie on the 2 circles.