Finding the point on a circle closest to a line in 3-space

Question

I've read that this problem is soluble generally for any circle and any line in 3D space, although I haven't found the solution itself.

However I'm interested in a special case where the circle lies in the $z = h$ plane for some constant $h$, and the centre of the circle is $(0,0,h)$. I'm hoping this might yield a simplified solution.

Suppose we have a straight line in Cartesian $xyz$-space such that $$ x = x_0 + v_x t, \quad \quad y = y_0 + v_y t, \quad \quad z = z_0 + v_z t$$

the circle in question is defined by $$ x = r_0 \cos{\phi}, \quad \quad y = r_0 \sin{\phi}, \quad \quad z = h$$

I can see that we could write down a function $D(\phi, t)$ for the distance between some point on the line and some point on the circle, but it's not clear to me how to minimise this function without resorting to a numeric solution.

Does an analytic solution exist in this case?

Thanks in advance!

Answer 1

1. This version of the problem is no simpler than the generic version, in the sense that the generic problem can, by a distance-preserving change of coordinates (translation plus orthogonal transformation), be changed into this one.

2. You can further simplify, by allowing scaling up or down uniformly, to the case where the circle is the unit circle in the $xy$-plane.

3. In that case, it's probably easiest to write the squared distance between a point $C$ of the circle, $C = (x, y, 0)$ with $x^2 + y^2 = 1$ and a point $S(t) = P + td$ of the line (where $P$ is a point on the line, and $d$ is a direction vector) and minimize that.

Let's look at that second problem. Instead of thinking of $C$ as being on the circle, let it be any point in space. What's the closest point $Q$ on the line that contains $P$ and has unit-vector direction-vector $d$? Answer: it's the one where $C-Q$ is perpendicular to $d$, so writing $Q = P + td$, we get $$ [(P + td) - C] \cdot d = 0 $$ which you can solve to get $$ t = \frac{(C-P) \cdot d}{d \cdot d} $$ which is just $(C-P) \cdot d$, because $d$ is a unit vector. So we can say $$ F(C) = P + ((C-P) \cdot d)d $$ is the point of the line closest to $C$. Note that this is linear in the coordinates of $C$.

Now restrict to points $(x, y)$ of the unit circle, parameterized, if you like, by $(\cos s, \sin s$. Then $F( (\cos s, \sin s) ) = L + M \cos s + N \sin s$ for some constant points/vectors $L, M, N$ that you are welcome to work out. You can compute the squared distance from this to $(\cos s, \sin s)$, which is then a quadratic in the cosine and sine. Take the derivative with respect to $s$ and set it to zero and solve, and you have found both the closest and farthest points.

In certain special cases (like the $z$-axis, passing through the middle of the unit circle in the xy plane, the coefficients $M$ and $N$ will both be zero vectors, so the quadratic will be a constant...this corresponds to all points on that circle being equidistant from the line.

This squared distance is

$$ U(C, t) = \|(P-C) + td\|^2 $$ The expression on the right, for a fixed $C$, is a quadratic in $t$ and has two solutions; with a little dancing around some special cases (like the line defined by $x = y = 0$, where there isn't a "point of the circle closest to the line", because all points are equally distance) you can actually pick the $t$ that gives the smaller squared-distance. (Every "raytracer" in computer graphics does this, but with a sphere instead of a circle). And from this $t$, you can find the closest point.

So now you have a function that takes an arbitrary point $C$ and produces the coordinates of the closest point on $L$.