= Solution
First assume $a\ne0$. The first equation is a <plane>, and the second is a <sphere> centred at $c$. Define its signed normal separation and projected centre by
$$
h=\frac{a\cdot(c-b)}{|a|},\qquad q=c-\frac{a\cdot(c-b)}{|a|^2}a.
$$
The <intersection of a plane and a sphere> is
$$
\boxed{\begin{cases}
\text{a circle with centre }q\text{ and radius }\sqrt{r^2-h^2},&|h|<r,\\
\{q\},&|h|=r,\\
\varnothing,&|h|>r.
\end{cases}}
$$
The <circle> lies in the given <plane>. Its radius follows from the <Pythagorean theorem>, splitting $x-c=(x-q)+(q-c)$ into <orthogonal> components. If $a=0$, the first equation is the identity $0=0$, so the locus is the \b[entire sphere], rather than a planar section.
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