Circle intersection in radial coordinates?

Question

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}

Answer 1

Using the identity:

\begin{align} \sin^2{\theta} + \cos^2{\theta} = 1 \end{align}

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 $\theta$.