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Radius first, circumscribed as the magic word solution in Creative category for Three Points Circle by maurice.makaay
from math import sqrt
from operator import add, sub
X, Y, VERTICAL = 0, 1, None
def checkio(data):
p1, p2, p3 = eval(data)
r = _radius_of_circumscribed_circle(p1, p2, p3)
x, y = _center_of_circumscribed_circle(p1, p2, p3, r)
return "(x-{0:g})^2+(y-{1:g})^2={2:g}^2".format(x, y, round(r, 2))
def _radius_of_circumscribed_circle(p1, p2, p3):
# Circumscribed circle: http://en.wikipedia.org/wiki/Circumscribed_circle
a = _distance(p1, p2)
b = _distance(p2, p3)
c = _distance(p3, p1)
area = _triange_area(a, b, c)
return a * b * c / (4 * area)
def _distance(p1, p2):
# Pythagoras: http://en.wikipedia.org/wiki/Pythagorean_theorem
a_square = (p1[X] - p2[X]) ** 2
b_square = (p1[Y] - p2[Y]) ** 2
return sqrt(a_square + b_square)
def _triange_area(a, b, c):
# Heron: http://en.wikipedia.org/wiki/Heron%27s_formula
s = (a + b + c) / 2;
return sqrt(s * (s-a) * (s-b) * (s-c));
def _center_of_circumscribed_circle(p1, p2, p3, radius):
# When drawing circles around two of the points using the provided radius,
# two intersection points can be found. One of these is the center of the
# circle that we're looking for (the circumcenter). By doing this for two
# sets of points, we can find the circumcenter (this one is in both the
# result sets).
i1 = set(_circle_intersections(p1, p2, radius))
i2 = set(_circle_intersections(p2, p3, radius))
return (i1 & i2).pop()
def _circle_intersections(p1, p2, r):
# http://mathforum.org/library/drmath/view/51836.html
(a, b), (c, d) = p1, p2
e = c - a
f = d - b
p = _distance(p1, p2)
k = p ** 2 / (2 * p)
for op in (add, sub):
x = op(a + e*k/p, (f/p) * sqrt(r**2 - k**2))
y = op(b + f*k/p, -(e/p) * sqrt(r**2 - k**2))
yield round(x, 2), round(y, 2)
Nov. 5, 2014
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