I recently had an exam and a question came up which I was only partially able to answer. The question was the following:
Let $\vec{a}$, $\vec{b}$ and $\vec{c}$ be constant vectors in $\mathbb{R}^{3}$ and let $r\in\mathbb{R}$. Find the locus of points for $\vec{x}$ resulting from the simultaneous solution of the equations: $$\vec{a}\cdot\vec{x}=\vec{a}\cdot\vec{b}$$ and $$\|\vec{x}-\vec{c}\|=r,$$ when: i) $r=0$, ii) $r>0$ iii) $r < 0$
The first one ($r=0$) is trivial, as we have from the second equation $\vec{x}=\vec{c}$, and therefore we can state that:
$$\vec{x}=\begin{cases}\{\vec{c}\} & \vec{b}=\vec{c} \lor \vec{a}=\vec{0} \\ \emptyset & \text{otherwise}\end{cases}$$
The third option $r<0$ again is trivial, we have from the second equation:
$$\|\vec{x}-\vec{c}\|=r <0 \implies \vec{x}=\emptyset$$
However, I am having trouble with part ii ($r>0$), we note that $\|\vec{x}-\vec{c}\|=r$ describes a sphere of radius $r$ about the point $\vec{c}$ and that $\vec{a}\cdot \vec{x}=\vec{a}\cdot \vec{b}$ describes a plane with normal $\vec{a}$ containing the point $\vec{b}$. With a bit of thought we can reduce this problem to two non-trivial cases; if we note that the perpendicular distance from the sphere to the plane is given by:
$$d=\left\|(\vec{b}-\vec{c})\cdot\frac{\vec{a}}{\|\vec{a}\|}\right\|$$
Therefore we have solution(s) for $\vec{x}$ iff:
$$r\geq \left\|(\vec{b}-\vec{c})\cdot\frac{\vec{a}}{\|\vec{a}\|}\right\|$$
If $r=d$ then we have that it touches the plane at the point:
$$\vec{x}=\vec{c}+\left((\vec{b}-\vec{c})\cdot \frac{\vec{a}}{\|\vec{a}\|^{2}}\right)\vec{a}$$
Else we have that it makes a circle in the plane orthogonal to $a$ centered at the point $\vec{c}+\left((\vec{b}-\vec{c})\cdot \frac{\vec{a}}{\|\vec{a}\|^{2}}\right)\vec{a}$ and with radius:
$$R=\sqrt{r^{2}-d^2}$$
However, I'm unable to come up with a simple form for $\vec{x}$ which satisfies this condition. I think the solution may involve writing the equation for the circle in $\hat{\boldsymbol{u}},\hat{\boldsymbol{v}},\hat{\boldsymbol{a}}$ co-ordinates and then converting it to $\hat{\boldsymbol{\imath}},\hat{\boldsymbol{\jmath}},\hat{\boldsymbol{k}}$, but I'm not sure, and all my attempts to do so have failed.
You don't need to find an expression fo $\vec {x}\,$:$\,$ it is enough to verify that $\|\vec{x}-\vec {c'}\|$ is constant
(with respect to $\vec{x}$) where $\vec {c'}$ is the (orthogonal) projection of $\vec {c}$ onto the plane.
If $d$ is the distance of $\vec {c}$ from the plane, then $$d=\frac {(\vec{b}-\vec{c})\cdot\vec{a}}{\|\vec{a}\|}$$ and $$\vec {c'}=\vec {c} + \frac d{\|\vec{a}\|}\vec{a}$$ So check yourself that$$(\vec{x}-\vec {c'})\cdot(\vec{x}-\vec {c'})=r^2-d^2$$ $\dots$ remembering that $$\vec{a}\cdot\vec{x}=\vec{a}\cdot\vec{b}$$$$$$See $\,$ W.A.Meyer $\,$ Geometry and Its Applications (2006), section 5.3, ex. 15, p. 241 .
Addendum
Of course the circle can be parameterized in a typical way.
Let $\,\vec {u}=\vec {b}-\vec {c'}$ and $\,\vec{v}=\vec {u} \times \vec {a}$ .
Then $$\vec {x}= \vec {c'}+ \sqrt {r^2-d^2} \left (\frac {\vec {u}}{\|\vec {u}\|} \cos t + \frac {\vec {v}}{\|\vec {v}\|} \sin t \right)$$ with $\,0 \le t < 2\pi$ .