Geometrically speaking, all points that are equidistance from points, p and q, all lie on the perdendicular bisector of pq.
So if $p = (x1, y1)$ and $q =(x2, y2)$ are the points, the line pq has slope, $\frac {y2 - y1}{x2 - x1}$. So the slope of a perpendicular line will have slope m = $- \frac {x2 - x1}{y2 - y1}$ As the line is a bisector, it will contain the midpoint $mid = (\frac{x1+x2}{2},\frac{y1+y2}{2})$. So the perpendicular bisector is the line with the equation $y - \frac{y1+y2}{2} = m(x - \frac{x1+x2}{2}) = y - \frac{y1+y2}{2} = - \frac {x2 - x1}{y2 - y1}(x - \frac{x1+x2}{2})$.
You want to find the points (x,y) where distance((x1,y1),(x,y)) = distance((x2,y2)(x,y)) = distance((x1,y1),(x2, y2)). Where (x,y) are on the line $y - \frac{y1+y2}{2} = - \frac {x2 - x1}{y2 - y1}(x - \frac{x1+x2}{2})$.
So you have three equations:
$y - \frac{y1+y2}{2} = - \frac {x2 - x1}{y2 - y1}(x - \frac{x1+x2}{2})$
$\sqrt{(x1 - x)^2 + (y1 - y)^2} = \sqrt{(x1 - x2)^2 + (y1 - y2)^2}$
and $\sqrt{(x2 - x)^2 + (y2 - y)^2} = \sqrt{(x1 - x2)^2 + (y1 - y2)^2}$
There will be two possible points that satisfy the the equations.
Okay, that's hard. BUT if you are given one of the two points the other will be symmetrically placed on the other side of the midpoint.
Ex: A=(7,3), B=(6,0), and C=(14,−1).
So the midpoint of A and B is M = (6.5, 1.5). C is 14 - 6.5 = 7.5 away from M in the x direction. C is -1 - 1.5 = -2.5 away in the y directions. So the 4th point X will also be 7.5 away in the x direction and -2.5 away in the y direction. So X = (6.5 - 7.5, 1.5 - (-2.5)) = (-1, 4).
See : https://en.wikipedia.org/wiki/Circumscribed_circle
In Cartesian system of coordinates :
$$\vec{P_1}=\left[\begin{matrix}x_1\\y_1\\z_1\end{matrix}\right]\qquad
\vec{P_2}=\left[\begin{matrix}x_2\\y_2\\z_2\end{matrix}\right]\qquad
\vec{P_3}=\left[\begin{matrix}x_3\\y_3\\z_3\end{matrix}\right]\qquad$$
The radius of the circle is :
$$R=\frac12\frac{\parallel\vec{P_1}-\vec{P_2}\parallel\:\parallel\vec{P_2}-\vec{P_3}\parallel\:\parallel\vec{P_3}-\vec{P_1}\parallel}{\parallel(\vec{P_1}-\vec{P_2})\times(\vec{P_2}-\vec{P_3})\parallel}$$
$\times\:$ is the cross product of vectors.
The center of the circle is given by :
$$\vec{P_c}=\alpha\:\vec{P_1}+\beta\:\vec{P_2}+\gamma\:\vec{P_3}$$
$$\alpha=\frac12\frac{\parallel\vec{P_2}-\vec{P_3}\parallel^2(\vec{P_1}-\vec{P_2})\bullet(\vec{P_1}-\vec{P_3})}{\parallel(\vec{P_1}-\vec{P_2})\bullet(\vec{P_2}-\vec{P_3})\parallel^2}$$
$$\beta=\frac12\frac{\parallel\vec{P_1}-\vec{P_3}\parallel^2(\vec{P_2}-\vec{P_1})\bullet(\vec{P_2}-\vec{P_3})}{\parallel(\vec{P_1}-\vec{P_2})\bullet(\vec{P_2}-\vec{P_3})\parallel^2}$$
$$\gamma=\frac12\frac{\parallel\vec{P_1}-\vec{P_2}\parallel^2(\vec{P_3}-\vec{P_1})\bullet(\vec{P_3}-\vec{P_2})}{\parallel(\vec{P_1}-\vec{P_2})\bullet(\vec{P_2}-\vec{P_3})\parallel^2}$$
$\bullet\:$ is the dot product of vectors.
Note for record : In case of a larger number of scattered points, a regression method is given in
https://fr.scribd.com/doc/31477970/Regressions-et-trajectoires-3D . This is also valid for three points only, but more complicated than the above method, thus less convenient as answer to the OP question.
Best Answer
Here's a picture for you: $$\color{blue}{‖x‖_1 = 1} \quad ‖x‖_2 = 1 \quad \color{orange}{‖x‖_5 = 1} \quad \color{red}{‖x‖_∞ = 1}$$
(It is sufficient to consider the unit balls centered at 0 because it "doesn't matter" where the centre is.) To understand this diagram, pick any point on the corresponding ball; that point is of $p$-distance $1$ from $0$.
As far as I know, the $p$-norms for $p=1,2,∞$ are the only cases that have a clear interpretation. Everything else is an 'interpolation' in between; if $p≈ 1$, then its the $1$-norm, but fudged slightly. If $p\gg 1$ then its like the $∞$-norm. As you can see, $p=5$ is already pretty close to the $∞$-norm.