Do two given circles intersect in Zero, One, or Two points and provide the intersection(s). The Stafford method may provide some guidance and alternate solution method. I will elaborate a more geometric solution utilizing Matlab specific functions, rotation matrix, and translation matrix. Assumption is that Matlab function circcirc is not available.
Given circles [x1,y1,R] and [x2,y2,P] return the intersections [], [x y], or [x y;x y].
The below figure is created based upon d=distance([x1,y1],[x2,y2]), translating (x1,y1) to (0,0), and rotating (x2,y2) to be on the Y-axis. From this manipulation two right triangles are apparent: [X,Y,R] and [X,d-Y,P]. Subtracting and simplifying these triangles leads to Y and two X values after substituting back into R^2=X^+Y^2 equation.
P^2=X^2+(d-Y)^2 and R^2=X^2+Y^2 after subtraction gives R^2-P^2=Y^2-(d-Y)^2 = Y^2-d^2+2dY-Y^2=2dY-d^2 thus
Y=(R^2-P^2+d^2)/(2d) and X=+/- (R^2-Y^2)^.5
The trick is to now un-rotate and translate this solution matrix using t=atan2(dx,dy), [cos(t) -sin(t);sin(t) cos(t)] and [x1 y1]
Diagram showing a normalization of posed problem where (x1,y1) is placed at origin and (x2,y2) is placed on Y-axis at distance d. Final (x,y) values will require rotation and shifting.
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Test cases updated 8/11/23 to include the top single point case. Revised template and my solution. Re-scoring not activated. The function norm is nice and sloow. I like the old days when scoring could be based on time and other creator functions.