How can I plot x2 + y2 = 9 as a circle not 3d equation in python

Question

How can I plot x^2 + y^2 = 9 as a circle not 3d equation in python

I cannot find any where to plot equality equation in python. Can you please help me out in this.

There seem to be multible functions to plot circles with for example matplotlib. As I do not know the context of your questions here the link to different methods: https://www.pythonpool.com/matplotlib-circle/ — Sam, Oct 27 '21 at 06:15

score 2 · Answer 1 · answered Oct 27 '21 at 06:15

2

You can use sympy as follows:

from sympy import Eq, plot_implicit
from sympy.abc import x, y

plot_implicit(Eq(x ** 2 + y ** 2, 9), aspect_ratio=(1, 1))

There is also plot_parametric:

from sympy import plot_parametric, cos, sin, pi
from sympy.abc import t

plot_parametric(cos(t), sin(t), (t, 0, 2 * pi), aspect_ratio=(1, 1))

answered Oct 27 '21 at 06:15

JohanC

is there anything similar for 3d plot – Subodh Dhabalia Oct 27 '21 at 09:20
There are `plot3d_parametric_line()` and `plot3d_parametric_surface()`, see the [sympy plotting docs](https://docs.sympy.org/latest/modules/plotting.html) for usage and examples. – JohanC Oct 27 '21 at 09:42

score 0 · Answer 2 · answered Oct 27 '21 at 06:17

You can derive x,y by solving the equation. For example, noting that -3 <= x <= 3 and y = +- sqrt(9-x^2),

import numpy as np

x = np.linspace(-3, 3)
y = np.sqrt(9-x)

import matplotlib.pyplot as plt
plt.scatter(x, y)
plt.scatter(x, -y)

Alternatively, since this relation can be expressed by the polar coordinate 0 <= theta < 2*pi with x = 3 cos(theta), y = 3 sin(theta),

theta = np.linspace(0, 2*np.pi)
x = 3*np.cos(theta)
y = 3*np.sin(theta)

plt.scatter(x, y)

How can I plot x**2 + y**2 = 9 as a circle not 3d equation in python