Talk:Circles of given radius through two points

More special cases

There may be more special cases. If p1==p2 and r==0, there is one unique answere that's a zero radius circle. If tow points are separated by exactly double the radius, there's only one answer. The latter can be treated as two identical circles, but then so can the former. <lang python>from math import sqrt

def find_center(p1, p2, r):

   if p1 == p2:
       if r == 0: return [p1] # special special case
       # maybe we can return a generator that yields random circles, eh?
       raise ValueError("infinite many answers")

   (x1,y1), (x2,y2) = p1, p2
   x, y = (x1 + x2)/2.0, (y1 + y2)/2.0
   dx, dy = x1 - x, y1 - y
   a = r*r / (dx*dx + dy*dy) - 1

   if not a: return [(x0, y0)]
   if a < 0: return []
   a = sqrt(a)
   return [(x + a*dy, y - a*dx), (x - a*dy, y + a*dx)]

print find_center((0, 0), (1, 1), 1) # normal case print find_center((0, 0), (0, 0), 0) # special case 1 print find_center((0, 0), (0, 2), 1) # special case 2</lang>

Hi Ledrug. I have one of those covered - two points on a diameter is tested by the current second set of inputs. I'll have to adjust for the two coincident points with r == 0.0 case. Thanks.

Hmm r==0.0 might be treated as an exception too as it is the circle as a point, (If you don't want points). --Paddy3118 (talk) 05:02, 17 April 2013 (UTC)

A circle with zero radius is still a perfectly valid circle, I don't see why it should be excluded. --Ledrug (talk) 19:55, 17 April 2013 (UTC)

But it does require the points to be coincident or it has no solution. –Donal Fellows (talk) 20:28, 17 April 2013 (UTC)

I was talking about when you do get a zero circle as a solution. --Ledrug (talk) 22:59, 17 April 2013 (UTC)

I could change it or leave it. Either case would work as the task explicitly states what to do with the r == 0.0 case, currently. --Paddy3118 (talk) 21:11, 17 April 2013 (UTC)

And dealing with nearly "normal" specifications are an important part of programming ;-)
--Paddy3118 (talk) 21:15, 17 April 2013 (UTC)