how to find center/radius of a sphere

Say you have an irregular tetrahedron, but you know the (x,y,z) coordinates of the four vertices; is there a simple formula for finding a sphere whose center exists within the tetrahedron formed by the four points and on whose surface the four points lie?

Simple formula, maybe not.

Take any three out of four points. The sphere in question must contain the circle through the three points within the plane of the three points. Which is to say, take three points, circumscribe a circle around that triangle. That circle has a center. If you draw a line through that center, orthogonal to that plane, the center of the sphere is somewhere along that line.

Now take out one of the points and put in the fourth, you now have a different three points in a different plane. Do the same things. The two lines must intersect in the center of the sphere.

If any three of the points are nearly collinear, or the four are nearly coplanar, there are any number of ways to make the method more robust

The other answers reduce to the two-dimensional problem.
Why not reduce it to the one-dimensional problem: The circumcentre is equidistant from each point.
Take the plane that is equidistant from $a$ and $b$, which is $2x\cdot(a-b)=|a|^2-|b|^2$.
There are six of these planes, take any three that involve all four points, and solve three equations in the three components of $x$