【Checkio Exercise】Three Point Circle

计算三角形外接圆的函数：

Three Point Circle

If we want to build new silos, then we need to make more formal and useful plots of land. Our topographer only marks a few points on the map and thinks that is good enough. It doesn't give us the coordinates which all our robots use, just a handwritten note. Paper notes! The nerve.

Through any three points that do not exist on the same line, there lies a unique circle. The points of this circle are represented in a string with the coordinates like so:

(x1,y1),(x2,y2),(x3,y3)

Where x1,y1,x2,y2,x3,y3 are digits.

You should find the circle for the three given points, such that the circle lies through these point and return the result as a string with the equation of the circle. In a Cartesian coordinate system (with an X and Y axis), the circle with central coordinates of (x0,y0) and radius of r can be described with the following equation:

(x-x0)^2+(y-y0)^2=r^2

Where x0, y0, r are decimal numbers rounded to two decimal points. Remove extraneous zeros and all decimal points, they are not necessary. For rounding, use the standard mathematical rules.

Input: Coordinates as a string.

Output: The equation of the circle as a string.

#答案
def circle_equation(note):
    note = list(eval(note))
    x1 = note[0][0]
    x2 = note[1][0]
    x3 = note[2][0]
    y1 = note[0][1]
    y2 = note[1][1]
    y3 = note[2][1]
    a = ((x3-x2)**2+(y3-y2)**2)**0.5
    b = ((x3-x1)**2+(y3-y1)**2)**0.5
    c = ((x1-x2)**2+(y1-y2)**2)**0.5
    p = 0.5*(a+b+c)
    s = (p*(p-a)*(p-b)*(p-c))**0.5
    x = ((y2 - y1) * (y3 * y3 - y1 * y1 + x3 * x3 - x1 * x1) - (y3 - y1) * (y2 * y2 - y1 * y1 + x2 * x2 - x1 * x1)) / (
2 * (x3 - x1) * (y2 - y1) - 2 * ((x2 - x1) * (y3 - y1)));
    y = ((x2 - x1) * (x3 * x3 - x1 * x1 + y3 * y3 - y1 * y1) - (x3 - x1) * (x2 * x2 - x1 * x1 + y2 * y2 - y1 * y1)) / (
2 * (y3 - y1) * (x2 - x1) - 2 * ((y2 - y1) * (x3 - x1)));
    r = a*b*c/(4*s)
    r = round(r, 2)
    if x.is_integer():
        x = int(x)
    if y.is_integer():
        y = int(y)
    if r.is_integer():
        r = int(r)
    string = '(x-' + str(x) + ')^2+(y-' + str(y) + ')^2=' + str(r) + '^2'
    return string

#调试结果
print(circle_equation("(2,2),(6,2),(2,6)"))
print(circle_equation("(3,7),(6,9),(9,7)"))
print(circle_equation("(2,2),(6,2),(2,6)") == "(x-4)^2+(y-4)^2=2.83^2")
print(circle_equation("(3,7),(6,9),(9,7)") == "(x-6)^2+(y-5.75)^2=3.25^2")