I would begin by imposing one circle centered at the origin of an \(xy\)-plane, and another centered at the point \((r,0)\), where \(2R<r\). Both circles have a radius of \(R\).

Now, let's find the line that passes through the point \(\displaystyle \left(\frac{r}{2},0\right)\), and which is tangent to the left circle. By symmetry, it will be tangent to the other circle as well. The line may be given as:

\(\displaystyle y=m\left(x-\frac{r}{2}\right)\)

Substituting this into the equation for the left circle, we obtain:

\(\displaystyle x^2+\left(m\left(x-\frac{r}{2}\right)\right)^2=R^2\)

Arrange this in standard quadratic form:

\(\displaystyle 4(m^2+1)x^2-4m^2rx+(m^2r^2-4R^2)=0\)

Equate the discriminant to zero:

\(\displaystyle (-4m^2r)^2-4(4(m^2+1))(m^2r^2-4R^2)=0\)

\(\displaystyle (m^2r)^2-(m^2+1)(m^2r^2-4R^2)=0\)

\(\displaystyle m^4r^2-m^4r^2+4m^2R^2-m^2r^2+4R^2=0\)

\(\displaystyle (4R^2-r^2)m^2+4R^2=0\)

\(\displaystyle m^2=\frac{4R^2}{r^2-4R^2}\)

Let's take the positive root:

\(\displaystyle m=\frac{2R}{\sqrt{r^2-4R^2}}\)

And so, our tangent line is given by:

\(\displaystyle y=\frac{2R}{\sqrt{r^2-4R^2}}\left(x-\frac{r}{2}\right)\)

Now, we are interested in the slope of a line perpendicular to this tangent line:

\(\displaystyle m_1=-\frac{\sqrt{r^2-4R^2}}{2R}\)

And so, the angle we desire will be:

\(\displaystyle \theta=2\arctan\left(\frac{\sqrt{r^2-4R^2}}{2R}\right)\)