Given three points $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$, if $$\frac{y_2-y_1}{x_2-x_1} \neq \frac{y_3-y_2}{x_3-x_2} \neq \frac{y_1-y_3}{x_1-x_3},$$ 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 $x = x_i$ and $y = y_i$, there will be a system of equation in three variables, that is:
\begin{align*}
\begin{pmatrix}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1 \\
\end{pmatrix}
\begin{pmatrix}
d \\
e \\
f
\end{pmatrix} &=
\begin{pmatrix}
-\left(x_1^2+y_1^2\right) \\
-\left(x_2^2+y_2^2\right) \\
-\left(x_3^2+y_3^2\right)
\end{pmatrix}.
\end{align*}
Solving this system gives the solution
\begin{align*}
d &= \frac{(x_3^2 + y_3^2 -x_1^2+y_1^2)}{x_1 – x_3} – e\left(\frac{y_1 – y_3}{x_1 – x_3}\right) \\
e &= \frac{(x_3^2 + y_3^2 -x_2^2+y_2^2)(x_1-x_3) – (x_3^2 + y_3^2 -x_1^2+y_1^2)(x_2-x_3)}{(y_2-y_3)(x_1-x_3) – (y_1-y_3)(x_2-x_3)} \\
f &= \frac{-(x_3^2 + y_3^2)(x_1-x_3) – (y_1-y_3)x_3}{x_1 – x_3} – e\left(\frac{y_3(x_1 – x_3) – x_3(y_1 – y_3)}{x_1 – x_3}\right)
\end{align*}
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?
Best Answer
Move the circle so that it passes through the origin (subtract a point from all three). The equation loses a coefficient,
$$x^2+y^2+dx+ey=0$$
which is easier to solve. Then reverse-translate.
$$-\frac d2=\dfrac{\begin{vmatrix}x_1^2+y_1^2&y_1\\x_2^2+y_2^2&y_2\end{vmatrix}}{2\begin{vmatrix}x_1&y_1\\x_2&y_2\end{vmatrix}}$$
$$-\frac e2= \dfrac{\begin{vmatrix}x_1&x_1^2+y_1^2\\x_2&x_2^2+y_2^2\end{vmatrix}}{2\begin{vmatrix}x_1&y_1\\x_2&y_2\end{vmatrix}}$$
are the coordinates of the center in the moved position.