finding a minimum distance

Question

I need to find a point or points on the given circle (or curve) which minimizes d0+d1? the radius and center of the curve are (0,0) and 'r' respectively and the coordinates of points A and B are known. Let say A=(x1,y1) and B=(x1,-y1) and r> sqrt(x1^2+y1^2) . C is unknown point of the circle which should minimize the length d0+d1 d0 - the distance between A to C on the circle d1- the distance between B to C on the circle

point C moves along the circle. I need to find a point or points on the given circle (or curve) which minimizes d0+d1?

Answer 1

The general case is very complicated, but the special situation

A=(x1,y1) and B=(x1,-y1) and r > sqrt(x1^2+y1^2)

with a circle whose centre is the origin has enough symmetries to make the solution at least in some circumstances accessible. I'm assuming A ≠ B, (equivalently y1 ≠ 0), otherwise the problem is trivial for a circle.

Let dist(P,Q) be the Euclidean distance between the points P and Q. The (closed) line segment connecting A and B is the locus of points P with

dist(P,A) + dist(P,B) = dist(A,B)

For D > dist(A,B), the locus of points with

f(P) = dist(P,A) + dist(P,B) = D

is an ellipse E(D) whose foci are A and B. Let P be a point on the circle and D = f(P).

• If the tangents to the circle and to the ellipse E(D) in the point P don't coincide, P is neither a local minimum nor a local maximum of f restricted to the circle.
• If the tangents coincide, and the curvature of the circle is larger than the curvature of E(D) in P, then P is an isolated local maximum of f restricted to the circle.
• If the tangents coincide, and the curvature of the circle is smaller than the curvature of E(D) in P, then P is an isolated local minimum of f restricted to the circle.
• If the tangents coincide and the curvature of the circle is equal to the curvature of E(D) in P, then
  • P is an isolated local minimum of f restricted to the circle if dist(P,A) = dist(P,B),
  • P is neither a local maximum nor a local minimum of f restricted to the circle otherwise.

First, if x1 = 0, it is easily seen (in case it is not geometrically obvious) that the points on the circle minimising f are the points with x-coordinate 0, i.e. P1 = (0,r) and P2 = (0,-r). [That would even be true if r² ≤ x1² + y1².]

Now, suppose x1 ≠ 0, without loss of generality x1 > 0. Then it is obvious that a point P = (x,y) on the circle minimising f must have x > x1. By the symmetry of the situation, the point R = (r,0) must either be a local minimum or a local maximum of f restricted to the circle.

Computing the behaviour of f near R, one finds that R is a local minimum if and only if

r ≥ (x1² + y1²) / x1

Since R is a point of smallest curvature of E(f(R)) (and the tangents in R to E(f(R)) and the circle coincide), R is then also the global minimum.

If r < (x1² + y1²) / x1, then R is a local maximum of f restricted to the circle. Then f has two global minima on the circle, with the same x-coordinate. Unfortunately, I don't have a nice formula to compute them, so I can't offer a better way than an iterative search.

Answer 2

If the line AB intersects the circle, then C is that intersection point (note that there can be two intersection points and both give an equal distance d0+d1 !).

If AB does not intersect the circle, then C is the point on the circle intersecting an imaginary line from the point on the line AB closest to the circle center.

There are many articles online about how to find the point on a line closest to another point, and how to find the intersection between two lines, which would solve the second case. For the first case you can google "line circle intersection"