Cutting in a circle/ellipse (PIL)

[flaerpen]

Hello, I wonder how to 'cut' or erase some parts of a image/circle that i've done in PIL.

Let say that I have two circles that intersects with eachother, I know the two points where they intersect and now I want to cut/erase the circle thats left.

Let me show you a image how I want it to look like, that I made in paint:

I really hope you understand and if not, tell me! /flaerpen

[flaerpen]

I have a possible answer. Instead of writing a ellipse/circle you write an Chord(?). but how do I use my points with the chord-function in PIL?

from http://www.pythonware.com/library/pi.../imagedraw.htm

chord

draw.chord(xy, start, end, options)

Same as arc, but connects the end points with a straight line.

The outline option gives the colour to use for the chord outline. The fill option gives the colour to use for the chord interior.

EDIT: Or maybe this is not the solution, because if there is more than one intersection with another circle how do you do that?

[flaerpen]

I now know (i think) that i'll use the chord-function, but have to ask how to calculate the start and end angle in a general formula.

I made a picture so it can be easier to show you what I really want:

Is there anyone who can help me to create a general forumula for the two green lines?

/flaerpen

[bvdet]

I now know (i think) that i'll use the chord-function, but have to ask how to calculate the start and end angle in a general formula.

I made a picture so it can be easier to show you what I really want:

Is there anyone who can help me to create a general forumula for the two green lines?

/flaerpen

flaerpen - This class will calculate the angles in radians with respect to circle 1 center point and the circle intersection points:

class CircleCircleIntersection(object):
    def __init__(self, p1, p2, r1, r2):
        '''
        Given circle center points p1 and p2 and circle radii r1 and r2
        Calculate intersection points of circles
        This only works in the X-Y plane at the present time
        '''
        self.p1 = p1
        self.p2 = p2
        self.r1 = r1
        self.r2 = r2
        self.d = p1.dist(p2)
        if self.d > r1+r2:
            self.Pa = None
            self.Pb = None
        elif self.d < abs(r1-r2):
            self.Pa = None
            self.Pb = None
        else:
            self.a = (r1**2-r2**2+self.d**2)/(2*self.d)
            self.b = self.d-self.a
            self.P0 = polarPt(p1, p2, self.a, p1)
            self.h = (r1**2-self.a**2)**0.5
            self.intPts()
 
    def intPts(self):
        self.Pa = Point()
        self.Pb = Point()
        self.Pa.x = self.P0.x + (self.h * (self.p2.y - self.p1.y) / self.d)
        self.Pb.x = self.P0.x - (self.h * (self.p2.y - self.p1.y) / self.d)
        self.Pa.y = self.P0.y - (self.h * (self.p2.x - self.p1.x) / self.d)
        self.Pb.y = self.P0.y + (self.h * (self.p2.x - self.p1.x) / self.d)
        self.theta0 = math.atan2(self.p2.y+self.p1.y, self.p2.x+self.p1.x)
        self.theta1 = self.theta0 - math.atan2(self.h, self.a)
        self.theta2 = self.theta0 + math.atan2(self.h, self.a)

Here is function 'polarPt':

def polarPt (p1, p2, d, p3=False):
    '''
    Calculate the vector p1-->p2
    Translate distance 'd' parallel to vector p1-->p2 from point 'p3'
    Return the new point
    'p3' defaults to 'p2'
    >>> polarPt(p1,p2,10,p3)
    Point(4.651484, 24.128709, 8.825742)
    >>> polarPt(p1,p2,10)
    Point(10.651484, 19.128709, 7.825742)
    >>> 
    '''
    return (p3 or p2)+(p2-p1).uv()*d

You will need a point object with 'x' and 'y' attributes, a unit vector (uv()) method, a distance (dist()) method, and + - * overloads. The angles calculated are counter-clockwise.

>>> a = CircleCircleIntersection(Point(), Point(-2,5.5,0), 3.5, 3.0)
>>> a.theta0
1.9195673303788037
>>> a.theta1
1.5052297601267781
>>> a.theta2
2.3339049006308294
>>> a.h
1.4090427459286017
>>> a.a
3.2038412164378536
>>> a.P0
Point(-1.094891, 3.010949, 0.000000)
>>> a.Pa
Point(0.229319, 3.492479, 0.000000)
>>> a.Pb
Point(-2.419100, 2.529418, 0.000000)
>>> 

I used Paul Bourke's website for reference. LINK