What is the easier way to find the circle given three points?
Given three points $({x}_{1},{y}_{1}),({x}_{2},{y}_{2})$, and $({x}_{3},{y}_{3})$, if
$\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\ne \frac{{y}_{3}-{y}_{2}}{{x}_{3}-{x}_{2}}\ne \frac{{y}_{1}-{y}_{3}}{{x}_{1}-{x}_{3}},$
then there will be a circle passing through them. The general form of the circle is
${x}^{2}+{y}^{2}+dx+ey+f=0.$
By substituting $x={x}_{i}\text{}\text{and}\text{}y={y}_{i}$, there will be a system of equation in three variables, that is:
$\begin{array}{rl}\left(\begin{array}{ccc}{x}_{1}& {y}_{1}& 1\\ {x}_{2}& {y}_{2}& 1\\ {x}_{3}& {y}_{3}& 1\end{array}\right)\left(\begin{array}{c}d\\ e\\ f\end{array}\right)& =\left(\begin{array}{c}-({x}_{1}^{2}+{y}_{1}^{2})\\ -({x}_{2}^{2}+{y}_{2}^{2})\\ -({x}_{3}^{2}+{y}_{3}^{2})\end{array}\right).\end{array}$
As there are a lot of things going around, the solution is prone to errors. Maybe this solution also has an error.
Is there a better way to solve for the equation of the circle?
Fitting a ballistic trajectory to noisy data where both spacial and temporal domains observations are noisy
Fitting a curve to noisy data is somewhat trivial. However it generally assumes that data abscissa is fixed, and the error is computed on the ordinate.
In my setup, I have 3D spacial observations of ballistic trajectories (that I model with a simple parabola), but the observations time are also noisy.
Therefore, I have to estimate the initial position $y}_{0},{y}_{0},{z}_{0$ and initial speed $v}_{{x}_{0}},{v}_{{y}_{0}},{v}_{{z}_{0}$, based on 4D (noisy) observations $({X}_{i},{Y}_{i},{Z}_{i},{T}_{i}),i\in [0,N]$, such that they fit the following model:
$\{\begin{array}{rl}x\left(t\right)& ={x}_{0}+{v}_{{x}_{0}}t\\ y\left(t\right)& ={y}_{0}+{v}_{{y}_{0}}t\\ z\left(t\right)& ={z}_{0}+{v}_{{z}_{0}}t-\frac{g}{2}{t}^{2}\end{array}$
with t monotonically increasing with i.
I'm not sure how to formulate such optimization problem because I have 6 parameters to estimate, but also 4N variables with only 3N equations… My intuition tells me there's only one single parabola that minimizes the error (MSE for example), but I can't formulate the problem.