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I have 2 sets of 4 coordinates on a xy plane with 4 Radii for each set. For the 1st set all the Radii are same forming it a perfect circle. where as for the second set the 4 Radii are little different forming a deformed circle. Can someone help me on how to plot these 2 shapes on a graph

Set1 : xy coordinates : [(0,1),(1,0),(0,-1),(-1,0)], Radii[1,1,1,1]

Set2 : xy coordinates : [(0,1),(1.2,0),(0,-0.9),(-1,0)], Radii[1,1.2,0.9,1]

I want the shape of set2 to look like a deformed circle. So, please help me with the math equation and the python implementation of the same.

Thanks for your time.

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  • Maybe this can help you: stackoverflow.com/questions/47873759/… Commented Jun 11, 2020 at 7:07
  • I don't understand what the coordinates represent. Do you just plot a circle with the given radius at the specified coordinates? If not, can you draw and attach a picture that shows what you want to get? Commented Jun 11, 2020 at 7:11

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Adapted from How to fit a 2D ellipse to given points.enter image description here

This should be a good start.

import numpy as np import matplotlib.pyplot as plt # circle = np.array([(0,1),(1,0),(0,-1),(-1,0)]) circle = np.array([(0,1),(1.2,0),(0,-0.9),(-1,0)]) # Extract x coords and y coords of the ellipse as column vectors X = circle[:,0:1] Y = circle[:,1:] # Formulate and solve the least squares problem ||Ax - b ||^2 A = np.hstack([X**2, X * Y, Y**2, X, Y]) b = np.ones_like(X) x = np.linalg.lstsq(A, b, rcond=-1)[0].squeeze() # Print the equation of the ellipse in standard form print('The ellipse is given by {0:.3}x^2 + {1:.3}xy +{2:.3}y^2 +{3:.3}x + {4:.3}y = 1'.format(*x)) # Plot the circle plt.scatter(X, Y, label='Data Points') # Plot the least squares ellipse x_coord = np.linspace(-2,2,300) y_coord = np.linspace(-2,2,300) X_coord, Y_coord = np.meshgrid(x_coord, y_coord) Z_coord = x[0] * X_coord ** 2 + x[1] * X_coord * Y_coord + x[2] * Y_coord**2 + x[3] * X_coord + x[4] * Y_coord plt.contour(X_coord, Y_coord, Z_coord, levels=[1], colors=('r'), linewidths=2) plt.legend() plt.xlabel('X') plt.ylabel('Y') plt.show()
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