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I have a rectangular frame and a circle with center and radius randomly generated. The center is always located within the limits of the frame, as shown:

enter image description here

I need to estimate the area of the fraction of the circle that is located within the frame. Currently I employ a simple Monte Carlo estimate that works ok, but I'd like to compare this with an exact geometric estimation of this area.

Is there a library and/or method to do this? I'm open to pretty much anything that can be installed with conda or pip.

import numpy as np
import matplotlib.pyplot as plt

def circFrac(cx, cy, rad, x0, x1, y0, y1, N_tot=100000):
    """
    Use Monte Carlo to estimate the fraction of the area of a circle centered
    in (cx, cy) with a radius of 'rad', that is located within the frame given
    by the limits 'x0, x1, y0, y1'.
    """

    # Source: https://stackoverflow.com/a/50746409/1391441
    r = rad * np.sqrt(np.random.uniform(0., 1., N_tot))
    theta = np.random.uniform(0., 1., N_tot) * 2 * np.pi
    xr = cx + r * np.cos(theta)
    yr = cy + r * np.sin(theta)

    # Points within the circle that are within the frame.
    msk_xy = (xr > x0) & (xr < x1) & (yr > y0) & (yr < y1)

    # The area is the points within circle and frame over the points within
    # circle.
    return msk_xy.sum() / N_tot


for _ in range(10):

    # Random (x, y) limits of the frame
    x0, y0 = np.random.uniform(0., 500., 2)
    x1, y1 = np.random.uniform(500., 1000., 2)

    # Random center coordinates *always* within the frame
    cx = np.random.uniform(x0, x1)
    cy = np.random.uniform(y0, y1)
    # Random radius
    rad = np.random.uniform(10., 500)

    frac = circFrac(cx, cy, rad, x0, x1, y0, y1)

    plt.xlim(x0, x1)
    plt.ylim(y0, y1)
    circle = plt.Circle((cx, cy), rad, fill=False)
    plt.gca().add_artist(circle)
    plt.scatter(
        cx, cy, marker='x', c='r', label="({:.0f}, {:.0f}), r={:.0f}".format(
            cx, cy, rad))
    plt.legend()
    plt.title("Fraction of circle inside frame: {:.2f}".format(frac))
    plt.axes().set_aspect('equal')
    plt.show()
Gabriel
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    This might be a question better suited to the Math SE. My math-fu isn't as strong as it once was, but I have a hunch calculus would give us an exact answer. – Engineero Oct 28 '19 at 19:06
  • Yes, I'm sure it would but I was hoping for an off-the-shelf solution already implemented in some library. – Gabriel Oct 28 '19 at 19:11
  • +1 for posting to Math SE. I reckon it could be possible to solve geometrically by splitting the circle up into sectors and triangles, but there are a lot of different cases to account for. This is a very specific problem for a library to have a solution for. – pigi5 Oct 28 '19 at 19:12
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    https://stackoverflow.com/questions/622287/area-of-intersection-between-circle-and-rectangle may give you some ideas. Despite being tagged Java, the top answer appears to be language-agnostic. – Kevin Oct 28 '19 at 19:15
  • Thank you Kevin (sorry for the very late answer!) I'll check those methods and try to come up with a Python implementation. – Gabriel Nov 04 '19 at 21:04

1 Answers1

2

You can use shapely for that:

import shapely.geometry as g

xr,yr,r = 0,0,5
circle = g.Point(xr,yr).buffer(r)

x1,y1,x2,y2 = -1,-2,3,5
rectangle = g.Polygon([(x1,y1),(x1,y2),(x2,y2),(x2,y1)])

intersection = rectangle.intersection(circle)
intersection.area

enter image description here

Vasily Mitch
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