On 03/20/2016 11:54 PM, Arlo Barnes wrote:
Anyway, now I realised I wanted something else. Given three points ABC, in either a closed or open triangle, I want a circle for which the two sides connected to the 'middle' node (they are ordered, remember) are both tangent(1) to said circle: tangent at the first(2) and(4) last(3) nodes, to be specific.
Hi,
Your triangle will have to be specific (isosceles, probably) for that circle to exist.
(1:tangent to AB and BC lines) <=> center is on angle bisector, radius is k|OB| for some angle-dependent constant k (it's too late for trigonomerty) (2: tangent to AB in A) <=> center is on perpendicular to AB going by A, radius is |OA| (3: tangent to BC in C) <=> center is on perpendicular to CB going by C, radius is |OC| (4: going through A and C) <=> center is on the median of AC, radius is |OA|=|OC|
those 4 lines have no reason to intersect if the triangle is not isosceles, so choose any compatible 2 (1&2,1&3,2&4,3&4)