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Category:2010s Hindi-language filmsQ:

Solving (or approximating) the circle equation in a matrix form?

$$x^2+y^2=r^2\tag{1}$$
Solving (1) is trivial, but is there a way to handle it in a matrix form?
What I’m looking for is:
$$\textbf{X}=\begin{bmatrix}x_1\\y_1\end{bmatrix},\begin{bmatrix}x_2\\y_2\end{bmatrix},\dots,\begin{bmatrix}x_n\\y_n\end{bmatrix}$$
where $\begin{bmatrix}x_i\\y_i\end{bmatrix}$ is the vector which contains the possible positions of the matrix and
$$\textbf{X}=\begin{bmatrix}\textbf{X}_1&\textbf{X}_2&\dots&\textbf{X}_n\end{bmatrix}\tag{2}$$
where $\textbf{X}_i$ is the matrix corresponding to the equation at position $i$.
If possible, I would like the matrix to give the same answer as the original vector.

A:

If you plot $x^2+y^2$ against $x$ at a constant $y$, you will see that when $y=0$, the equation $x^2+y^2=r^2$ means $x=r$. So, what we are doing is, we are plotting $x^2+y^2$ against $x$ for a constant $y=0$.
Now, if $x_1, x_2, x_3$ are the possible values of $x$ (for $y=0$), then what we do is, plot $x^2_1+y^2_1$, $x^2_2+y^2_2$, $x^2_3+y^2_3$ against \$x

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