We have two circles in the plane described by $C_0 = (x_0, y_0, r_0)$ and $C_1 = (x_1, y_1, r_1)$

We know that they intersect but one does not completely overlap the other. That is to say their interiors are neither disjoint nor is one a subset of the other.

Clearly the borders of the two circles intersect at exactly two points.

If we describe the points of $C_0$ in "parametric radial coordinates" as:

\begin{align} P(\theta) = (x_0 + r_0\cos{\theta}, y_0 + r_0\sin{\theta}) \end{align}

Then there are two values of $\theta \in [0,2\pi)$ corresponding to the two border intersection points such that:

\begin{align} r_1 &= |P(\theta) - (x_1,y_1)| \\ {r_1}^2 &= (x_0 - x_1 + r_0\cos{\theta})^2 + (y_0 - y_1 + r_0\sin{\theta})^2 \tag{1} \end{align}

How do I solve eq.(1) for $\theta$ ?

If I assign $d_x = x_0 - x_1$ and $d_y = y_0 - y_1$ and expand the rhs I get:

\begin{align} {r_1}^2 = {d_x}^2 + 2d_xr_0\cos{\theta} + {r_0}^2\cos^2{\theta} + {d_y}^2 + 2d_yr_0\sin{\theta} + {r_0}^2\sin^2{\theta} \end{align}

But then I am equally stuck.

\begin{align} \theta = {???} \end{align}

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You must also consider the case where the circles are tangent to each other, i.e., they intersection in only one point (such as the unit circle centered at $(0,0)$ and another circle of radius $1$ centered at $(2,0)$). – Clayton Jan 11 '13 at 3:33
@Clayton: Then they would not qualify as their interiors would be disjoint. See second sentence of post. – Andrew Tomazos Jan 11 '13 at 3:34
So then the condition is stronger than saying the circles intersect; that was what I wanted to know. – Clayton Jan 11 '13 at 3:36
Why not work with rectangular coordinates? – leo Jan 11 '13 at 3:45
@leo: I specifically need the $\theta$ for the application. But actually you are right - I could solve in rectangular coordinates and then calculate theta after having the two points (x,y) coordinates. – Andrew Tomazos Jan 11 '13 at 3:47

You can substitute an expression involving $\sin$ for $\cos$ and visa verse.
By substituting this identity in your expanded expression you can solve for $\sin{\theta}$ and hence for $\theta$.