I have a set of points on a plane, and I want to join these points using a circular arc between consecutive points such that the final curve I get is smooth (no sharp edges).

Is this possible? If so, how can I find that arc?

  • $\begingroup$ To proceed with the circular arc attempt, you would find a set of circles such that each circle has (at least) two points from the given set on its edge, and such that every circle is tangent to the circle "next" to it, i.e., two circles which are to connect to each other will do so in only one point. $\endgroup$
    – abiessu
    Mar 11, 2015 at 13:00

1 Answer 1


First I will rewrite the problem in a more precise terms.

Given $n>1$ points in the plane, $A_1$, $\ldots$, $A_n$, prove or disprove the existence of $n-1$ circular arcs $A_1A_2$, $\ldots$, $A_{n-1}A_n$ such that $A_{k-1}A_k$ is tangent to $A_kA_{k+1}$ for every $2\le k\le n-1$.

Draw any arc between $A_1$ and $A_2$ (the center $C_1$ must be on the perpendicular bisector of $\overline{A_1A_2}$). Now, draw the line $C_1A_2$. This line shall intersect the perpendicular bisector of $A_2A_3$ at $C_2$. This is the center of the second arc. If $C_1A_2$ and the perpendicular bisector of $A_2A_3$ are parallel, $C_2$ will be the middle point of $A_2A_3$.

Can you continue?

  • $\begingroup$ I forgot... this is related to the work of Voronoi, right? $\endgroup$
    – abiessu
    Mar 11, 2015 at 13:03
  • $\begingroup$ This is surely helpful :) . A further optimisation I wanted to add was to minimise the curvature of the curve. I think I can find some $C_1$ to do that $\endgroup$
    – udiboy1209
    Mar 11, 2015 at 13:07
  • $\begingroup$ @udiboy1209 If you want to allow for minimal curvature, you need to also allow for the drawing of straight lines, because for two (or more) colinear points, you can draw a set of circular arcs that have an arbitrarily small curvature, but you cannot achieve zero curvature. $\endgroup$
    – 5xum
    Mar 11, 2015 at 13:10

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