Given two points and a an angle, find the radius of the circle that intersects both points

Question

Given the points a and b and the angle alpha (which is the angle between the vector from a to b and the circle tangent vector at a), how do I find the radius r of the circle that intersects a and b?

Answer 1

I presume that you know the distance $AB$. Let's call $d$ this distance.

We also know all the angles in triangle ABC: \begin{align*} \hat{A} & = \frac{\pi}{2} - \alpha \\ \hat{B} & = \hat{A} = \frac{\pi}{2} - \alpha \\ \hat{C} & = \pi - \hat{A} - \hat{B} = \pi - 2 \left( \frac{\pi}{2} - \alpha \right) = 2 \alpha \\ \end{align*}

Then by application of Al Kashi's theorem to triangle ABC: \begin{align*} AB^2 & = AC^2 + BC^2 - \cos(\hat{C}) \\ d^2 & = 2 r^2 - \cos(\alpha) \\ r^2 & = \frac 1 2 \left( d^2 + \cos(2 \alpha) \right) \\ \end{align*}