Perhaps I didn't use the appropriate terms to describe my question, so I'll try my best to describe it.
Imagine a line in a 3D plane. Now imagine a circle forming around it at a certain point, where all the perpendiculars of the line intersect with the circle.
I need to be able to find the equation of the circle (relative to the point it formed around).
I know it's a broad question, but all of my searches didn't yield any results.
Denote the distinguished point by ${\bf x}_0 = (x_0, y_0, z_0)$, pick any (nonzero) vector ${\bf a} = (a, b, c)$ parallel to the line, and let $R$ denote the desired radius of the circle. Then, the circle is locus of points in the plane orthogonal to the line (and hence $\bf v$) at ${\bf x}_0$ that are a distance $R$ from ${\bf x}_0$, that is, the points $\bf x$ satisfying $${\bf a} \cdot ({\bf x} - {\bf x}_0) = 0 \qquad \text{and} \qquad ({\bf x} - {\bf x}_0) \cdot ({\bf x} - {\bf x}_0) = R^2.$$ In components this is $$a (x - x_0) + b (y - y_0) + c (z - z_0) = 0 \qquad \text{and} \qquad (x - x_0)^2 + (y - y_0)^2 + (z - z_0)^2 = R^2.$$
One can parameterize this circle by picking an orthonormal pair $({\bf p}, {\bf q})$ of vectors each orthogonal to ${\bf v}$ and declaring $$\gamma(t) := {\bf x}_0 + R \cos t \,{\bf p} + R \sin t \,{\bf q}.$$