 Solution

Subtract the two constraints to obtain $x\cdot (a-b)=A-B$. The Cauchy-Schwarz inequality gives $|A-B|\le \Vert x\Vert \Vert a-b\Vert$. Since $a\ne b$, division and squaring give the distance bound from two hyperplane constraintsIf $a=b$ and $A\ne B$, the intersection is empty. If $a=b\ne 0$ and $A=B$, the constraints describe the same hyperplane. Its points satisfy $\Vert x{\Vert }^2\ge A^2/\Vert a{\Vert }^2$, with equality at $x=Aa/\Vert a{\Vert }^2$. The fraction involving $a-b$ is undefined in this case and cannot be used.

If $a=b=0$, both equations are satisfied by every $x$ when $A=B=0$, and otherwise there is no solution. These degenerate cases distinguish an inconsistent constraint from a repeated one.