Here's a sketch of the problem:
Since AB1 is tangent to the circle and B1C is the radius of the circle, angle AB1C is a right angle. The distances f and from A to C, say, d, may be calculated using the pythagorean theorem. Now, we draw the altitude of triangle AB1C to E. By similar triangles (ΔAB1C ~ ΔB1EC~ ΔAEB1) we determine that
and
and so, e = r2/d and a = f r/d. We let δ be the normalized direction vector from C to A, namely,
Then point E is given by E = C + e δ. Let δp= (δy, –δx), or the clockwise 90° rotation of δ. Then points B1 and B2 are given by:
B1 = E + a δp
and
B2 = E – a δp,
which are the desired points.
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