I suggest turning this into a 2D problem and then find the circle from three points on the plane.
NOTE: Below the $\times$ is vector cross product and $\cdot$ the vector dot product. Also $^\top$ is a matrix transpose (switch rows with columns). Finally $\| \boldsymbol{x} \|$ returns the magnitude of the vector $\boldsymbol{x}$.
Start by finding the normal vector to the plane defined by the three points
$$ \boldsymbol{n} = \mathrm{unitvector}( \boldsymbol{A} \times \boldsymbol{B} + \boldsymbol{B} \times \boldsymbol{C} + \boldsymbol{C} \times \boldsymbol{A}) \tag{1}$$
The find the arbitrary mutually orthogonal directions
$$ \boldsymbol{u} = (\boldsymbol{C}-\boldsymbol{A})\times \boldsymbol{n} \tag{2}$$
$$ \boldsymbol{v} = \boldsymbol{n} \times \boldsymbol{u} \tag{3}$$
and construct a 3×2 rotation matrix from the three direction vectors as columns.
$$ \mathbf{R} = \left[ \boldsymbol{u} \; \boldsymbol{v} \right] \tag{4}$$
You also need the distance of the plane ABC to the origin
$$ d = \boldsymbol{n} \cdot \boldsymbol{A} \tag{5}$$
Now convert the problem into a 2D problem with
$$ \begin{aligned}
\boldsymbol{a} = \pmatrix{a_x\\a_y} & = \mathbf{R}^\top \boldsymbol{A}\\
\boldsymbol{b} = \pmatrix{b_x\\b_y} & = \mathbf{R}^\top \boldsymbol{B}\\
\boldsymbol{c} = \pmatrix{c_x\\c_y} & = \mathbf{R}^\top \boldsymbol{C}
\end{aligned} \tag{6}$$
Now solve the 2D problem by finding the center point $\boldsymbol{q}=\pmatrix{x\\y}$ using a 2×2 system of equations
$$ \begin{bmatrix} b_x-a_x & b_y -a_y \\ c_x -a_x & c_y - a_y \end{bmatrix} \pmatrix{x\\y} = \pmatrix{ \frac{ (b_x^2+b_y^2)-(a_x^2+a_y^2)}{2} \\ \frac{ (c_x^2+c_y^2)-(a_x^2+a_y^2)}{2} } \tag{7} $$
Lastly construct the 3D point for the circle center $\boldsymbol{Q}$
$$ \boldsymbol{Q} = \boldsymbol{n}\, d + \mathbf{R} \pmatrix{x\\y} \tag{8}$$
If the radius is needed just take the distance from the center to any point
$$ \mathrm{radius} = \| \boldsymbol{A}-\boldsymbol{Q} \| \tag{9} $$
You might also need the sweep angle which you get from the angle between the vectors $\boldsymbol{C}-\boldsymbol{Q}$ and $\boldsymbol{A}-\boldsymbol{Q}$
$$ \theta = \cos^{-1} \left( \frac{ (\boldsymbol{C}-\boldsymbol{Q}) \cdot ( \boldsymbol{A}-\boldsymbol{Q}) }{ \| \boldsymbol{C}-\boldsymbol{Q}\| \,\|\boldsymbol{A}-\boldsymbol{Q} \| } \right) \tag{10} $$
Example
Points:
$$\begin{aligned} \boldsymbol{A} &= \pmatrix{1\\0\\0} & \boldsymbol{B} & = \pmatrix{3\\1\\0} & \boldsymbol{C} &= \pmatrix{2 \\0 \\ -1} \end{aligned} $$
Normal:
$$ \boldsymbol{n} = \pmatrix{
-\tfrac{\sqrt{6}}{6} \\
\tfrac{\sqrt{6}}{3} \\
-\tfrac{\sqrt{6}}{6} } $$
Distance from Origin
$$ d = -\tfrac{\sqrt 6}{6} $$
Rotation:
$$ \mathbf{R} = \begin{bmatrix}
\tfrac{\sqrt 3}{3} & \tfrac{\sqrt 2}{2} \\
\tfrac{\sqrt 3}{3} & 0 \\
\tfrac{\sqrt 3}{3} & -\tfrac{\sqrt 2}{2}
\end{bmatrix} $$
2D Points:
$$ \begin{aligned}
\boldsymbol{a} & = \pmatrix{ \tfrac{\sqrt 3}{3} \\ \tfrac{\sqrt 2}{2} } &
\boldsymbol{b} & = \pmatrix{ \tfrac{4 \sqrt 3}{3} \\ \tfrac{3 \sqrt 2}{2} } &
\boldsymbol{c} & = \pmatrix{ \tfrac{\sqrt 3}{3} \\ \tfrac{3 \sqrt 2}{2} }
\end{aligned} $$
System of equations:
$$\left. \begin{bmatrix} \sqrt{3} & \sqrt{2} \\ 0 & \sqrt{2} \end{bmatrix} \pmatrix{x \\ y} = \pmatrix{\tfrac{9}{2} \\ 2 } \;\right\} \; \pmatrix{x = \tfrac{5 \sqrt 3}{6}\\y = \sqrt 2} $$
Circle Center:
$$ \boldsymbol{Q} = \left( -\tfrac{\sqrt 6}{6}\right) \pmatrix{
-\tfrac{\sqrt{6}}{6} \\
\tfrac{\sqrt{6}}{3} \\
-\tfrac{\sqrt{6}}{6} } + \begin{bmatrix}
\tfrac{\sqrt 3}{3} & \tfrac{\sqrt 2}{2} \\
\tfrac{\sqrt 3}{3} & 0 \\
\tfrac{\sqrt 3}{3} & -\tfrac{\sqrt 2}{2}
\end{bmatrix} \pmatrix{ \tfrac{5 \sqrt 3}{6}\\ \sqrt 2} = \pmatrix{2 \\ \tfrac{1}{2} \\ 0} $$
Prove that points A, B, and C are equidistant to Q
$$ \mathrm{radius} = \| \boldsymbol{A}-\boldsymbol{Q} \| = \| \boldsymbol{B}-\boldsymbol{Q} \| = \| \boldsymbol{C}-\boldsymbol{Q} \| = \tfrac{\sqrt 5}{2} \;\;\checkmark $$
PS. I am also a fellow roboticist. You do need a basic understanding of linear algebra operations as well as some vector calculus to understand the how and why of things in robotics.
