Find smallest circle that encloses jordan curve

Question

Let $\mathbf{p}(t)$ parametrize a jordan curve in the plane with the interval $[0, \ 1]$.

How can I find the smallest circle that encloses all points of this curve?

This question relates to the Smallest-circle problem

Examples:

1. If $\mathbf{p}$ describes the boundary of a polygon, the circle would be the one that contains all the vertices
2. If $\mathbf{p}$ describes the circle, then the bounding circle would be equal to $\mathbf{p}$.

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Basically I'm looking for an algorithm to find this circle.

I thought about using an algebraic approach: find three parameters $t_0$, $t_1$ and $t_2$ to maximize the radius over these three variables. From this question we get the expression

$$r = \left| \left(z_1-z_0\right) \cdot \dfrac{w-|w|^2}{w-\overline{w}}\right|; \ \ \ \ \ \ w = \dfrac{z_2-z_0}{z_1-z_0}$$

With $z_0$, $z_1$ and $z_2$ being the complex numbers generated by the position of $\mathbf{p}$ at values $t_0$, $t_1$, $t_2$

To find the maximum radius, I would use Newton's method with some starting values, but is there another way?

Example: Reduce the problem to one of finding the circle that passes through $n$ points, the candidates, then the problem is easier.

1. Set the candidates the points which $\mathbf{p}$ is not analitic, at the joints
2. Add the points with smallest curvature to the candidates
3. Compute the baricenter, move it to the origin and add the furtherest point.
4. Divide the curve by segments, find the bounding circle of each segment, and then find the biggest circle that encloses all the smaller circles

Notes:

1. $\mathbf{p}$ is limited
2. $\mathbf{p}$ is analitic by parts
3. $\mathbf{p}$ doesn't intersect itself
4. $\mathbf{p}$ is closed

Answer 1

For an algorithmic solution, which will be approximate, you can,

• flatten the curve(s), i.e. decompose it in successive segments that do not deviate from the curve by more than a specified distance;

• construct the bounding circle of the vertices; this can be done on all points, or only on the extreme points, obtained by a convex hull algorithm.

In some specific cases, such as when the curve is made of circular arcs, an exact solution is possible, but can be difficult. The above solution will work with any curve, with a controllable degree of accuracy.

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For a more analytical solution, you can consider the function that associates to a point $(x,y)$ the farthest distance to the [parametric] curve, obtained by maximizing

$$(x-u(t))^2+(y-v(t))^2$$ for $t$ in the definition range. The stationary points are the solutions of

$$(x-u(t))u'(t)+(y-v(t))v'(t)=0$$ and they correspond to extrema of the radius. The maximum maximorum must be retained. Now we have a function $r(x,y)$ and maximize it by canceling the gradient.

Alternatively, we can consider the minimization problem of

$$(x-u(t))^2+(y-v(t))^2$$ under the constraint $$(x-u(t))u'(t)+(y-v(t))v'(t)=0,$$ which can be done with Lagrange multipliers. Additional conditions are necessary to obtain the minmax solution.

E.g. consider the circle $u=\cos(t),v=\sin(t)$. After simplification, the maximum distance condition is $$x\sin(t)-y\cos(t)=0,$$ giving the expression of the squared radius

$$\left(\sqrt{x^2+y^2}+1\right)^2.$$

This function is clearly minimized at $x=0,y=0$, with $r=1$.

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There is also a parasitic solution corresponding to the minima of

$$\left(\sqrt{x^2+y^2}-1\right)^2,$$ which are on the circle itself and correspond the minima of the distance.

Answer 2

I haven't fully established this, but you might find it useful so I want to give you the idea sooner rather than later. I need to do some additional reading, but I think the approach is promising, and I will come back to edit this comment if this turns out to be justifiable.

Modified problem

Let $p$ be a piece-wise $C^1$, bounded, simple, closed curve in $\mathbb{R}^2$. Find the smallest enclosing circle for this curve.

Approach

If we find the largest distance between two points on this curve, we can use these two points to construct a line segment corresponding the diameter of the enclosing circle.

Perform a translation so that the origin lies in the interior of the curve.

Use Lagrange multipliers to find the points on the curve farthest from the origin, say $P$ and $Q$.

The midpoint $\bigg(\dfrac{P_x+Q_x}{2},\dfrac{P_y+Q_y}{2}\bigg)$ gives the center of the enclosing circle, and $\dfrac{||P-Q||}{2}$ gives the radius.

Apply the reverse translation on the center of this circle (e.g. if your first translation was $T(x,y)=(x+a,y+b)$, now apply $T^{-1}(x,y)=(x-a,y-b)$). This brings the center of the circle enclosing the translated curve to the same relative position with respect to the original curve.

Now draw a circle of radius $\dfrac{||P-Q||}{2}$ around that point. This circle encloses the original curve.