Overlapping circles

Question

Consider two circles of radii $R$ and $r$ whose centres are separated by a distance $d$.

The area of overlap between them (assuming $|R-r| \le d \le R+r$) is given by the following formula, derived below.

$$A(d; R, r) = \beta R^2 + \alpha r^2 - \frac{1}{2}r^2\sin 2\alpha - \frac{1}{2}R^2\sin 2\beta,$$ where $$\cos \alpha = \frac{r^2 + d^2 - R^2}{2rd} \quad \mathrm{and} \quad \cos \beta = \frac{R^2 + d^2 - r^2}{2Rd}.$$

Inverting this formula to find $d$ for a given overlap area $A=A_0$ cannot be achieved analytically, but a numerical solution can be found as the root of the function $A(d; R, r) - A_0$.

Write a Python function which takes arguments A, the target overlap area, and R and r the two circle radii and returns d, the distance between the circle centres giving overlap area A. Use, for example, brentq.

Derivation

First note that $A=0$ if $d \ge R+r$: the circles do not intersect at all in this case. Also, $A = \pi (\mathrm{min}(R,r))^2$ if $d \le |R-r|$: the smaller circle is entirely enclosed in the larger in this case. For the case of partial overlap, $|R-r| < d < R+r$ the area to find is shaded in the diagram below.

The cosine formula gives the angles $\alpha$ and $\beta$:

$$\cos \alpha = \frac{r^2 + d^2 - R^2}{2rd} \quad \mathrm{and} \quad \cos \beta = \frac{R^2 + d^2 - r^2}{2Rd}.$$

The required area may be found as the sum of the two circle segments cut off by the chord CD. For the circle centred at A in the diagram above, its segment is the area of the circular sector ACD minus the area of the triangle ACD:

With $x=\mathrm{AE} = R\cos\beta$ and $h=\mathrm{CE}=R\sin\beta$, triangle ACD has area $xh=R^2\cos\beta\sin\beta=\frac{1}{2}R^2\sin2\beta$. The area of the circular sector ACD is simply $\beta R^2$ with $\beta$ measured in radians (the entire circle has area $\pi R^2$ and we want the fraction $\beta/2\pi$ of it).

Therefore, the shaded circular segment has area $\beta R^2 - \frac{1}{2}R^2\sin2\beta$.

Similarly, the area of the segment of the circle centred at B cut off by chord CD is $\alpha r^2 - \frac{1}{2}r^2\sin2\alpha$.

The total intersection area is therefore

$$A = \alpha r^2 + \beta R^2 - \frac{1}{2}r^2\sin2\alpha - \frac{1}{2}R^2\sin2\beta.$$

Solution

d, R, r = float(d), float(R), float(r)

import numpy as np
from scipy.optimize import brentq

def intersection_area(d, R, r):
    """Return the area of intersection of two circles.

    The circles have radii R and r, and their centres are separated by d.

    """

    if d <= abs(R-r):
        # One circle is entirely enclosed in the other.
        return np.pi * min(R, r)**2
    if d >= r + R:
        # The circles don't overlap at all.
        return 0

    r2, R2, d2 = r**2, R**2, d**2
    alpha = np.arccos((d2 + r2 - R2) / (2*d*r))
    beta = np.arccos((d2 + R2 - r2) / (2*d*R))
    return ( r2 * alpha + R2 * beta -
             0.5 * (r2 * np.sin(2*alpha) + R2 * np.sin(2*beta))
           )

def find_d(A, R, r):
    """
    Find the distance between the centres of two circles giving overlap area A.

    """

    # A cannot be larger than the area of the smallest circle!
    if A > np.pi * min(r, R)**2:
        raise ValueError("Intersection area can't be larger than the area"
                         " of the smallest circle")
    if A == 0:
        # If the circles don't overlap, place them next to each other
        return R+r

    if A < 0:
        raise ValueError('Negative intersection area')

    def f(d, A, R, r):
        return intersection_area(d, R, r) - A

    a, b = abs(R-r), R+r
    d = brentq(f, a, b, args=(A, R, r))
    return d

In [x]: find_d(0.35, 1, 0.4))
0.8489711717559927