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 $$\frac{y_2-y_1}{x_2-x_1} \neq \frac{y_3-y_2}{x_3-x_2} \neq \frac{y_1-y_3}{x_1-x_3},$$ 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?

I prefer a different approach.
You know that the center of the circle is the intersection between any two of the three perpendicular bisectors.
Therefore,
$$(x - x_1)^2 + (y - y_1)^2 = (x - x_2)^2 + (y - y_2)^2 \tag1 $$
and
$$(x - x_1)^2 + (y - y_1)^2 = (x - x_3)^2 + (y - y_3)^2. \tag2 $$
Let $f_1(x) = (x - x_1)^2 - (x - x_2)^2. ~~: ~~$ 1st degree polynomial
Let $f_2(x) = (x - x_1)^2 - (x - x_3)^2. ~~: ~~$ 1st degree polynomial
Let $g_1(y) = (y - y_2)^2 - (y - y_1)^2. ~~: ~~$ 1st degree polynomial
Let $g_2(y) = (y - y_3)^2 - (y - y_1)^2. ~~: ~~$ 1st degree polynomial
You can express $g_2(y)$ as $Ag_1(y) + B$, where $A,B$ are fixed real numbers.
This implies that
$$f_2(x) = g_2(y) = Ag_1(y) + B = Af_1(x) + B. \tag3 $$
Equation (3) above gives you an equation solely in $x$, that you can solve to determine the $(x = x_0)$ coordinate of the center of the circle.  Once this is determined, you can compute $y_0$ via $g_1(y_0) = f_1(x_0).$
