One hypersphere at $x^2 + y^2 + z^2 + w^2 = 1$ intersects another hypersphere at $(x - 1)^2 + y^2 + z^2 + w^2 = 1$.
(EDIT to address comments)
The intersection results in a (3-d) sphere with radius $ \frac{\sqrt 3}{2}$ contained within the hyperplane $x = \frac{1}{2}$
Why?
I can see the derivation of $x = 1$ through simultaneously solving both equations. I assume the radius comes from using the Pythagorean theorem in multiple dimensions, but I just can't imagine it in order to derive it.
Any explanation would be appreciated.
How do you get $x = 1$ from that? If you take the differences between the two equations: $$\begin{array}{rrr}&x^2 + y^2 + z^2 + w^2 = 1\\ -[&(x-1)^2 + y^2 + z^2 + w^2 = 1&]\\\hline &x^2-(x-1)^2 = 0\\&2x-1 = 0\\&x=\frac 12\end{array}$$
And even without solving the equation, just geometrically, it is obvious by symmetry that the intersection of two hyperspheres of the same radius must lie in the hyperplane halfway between their centers, not a hyperplane passing through one of the centers.
As for the rest, once you realize the sphere has $x = \frac 12$, just substitute that value into one of the equations (both give the same result): $$\left(\frac 12\right)^2 + y^2 + z^2 + w^2 = 1\\ y^2 + z^2 + w^2 = \frac 34 = \left(\frac{\sqrt 3}2\right)^2$$
which is the equation of a sphere of radius $\frac{\sqrt 3}2$.