如何将缺失点绘制成一个完整的圆？

import numpy as np
from scipy.interpolate import interp1d

# 假设我们有一组x和y坐标，其中一些y值是缺失的（用np.nan表示）
x = np.array([1, 2, 3, np.nan, 5])
y = np.array([2, np.nan, 4, 5, 6])

# 使用线性插值填补缺失值
f = interp1d(x[~np.isnan(x)], y[~np.isnan(y)], kind='linear', fill_value="extrapolate")
x_filled = x.copy()
x_filled[np.isnan(y)] = f(np.where(np.isnan(y))[0])
y_filled = f(x_filled)

print("Filled x:", x_filled)
print("Filled y:", y_filled)

from scipy.optimize import least_squares

def circle_fit(x, y):
    def residuals(params, x, y):
        xc, yc, r = params
        return (x - xc)**2 + (y - yc)**2 - r**2

    initial_guess = [np.mean(x), np.mean(y), np.std(x)]
    result = least_squares(residuals, initial_guess, args=(x, y))
    return result.x

# 使用圆拟合算法找到圆的参数
circle_params = circle_fit(x_filled, y_filled)
print("Circle parameters (xc, yc, r):", circle_params)