How to find center of a circle from only an arbitary arc of that circle

How to find the center of a circle with given an arbitrary arc. we only have the arc nothing else. Is there any known equation or way to complete the circle.

Solution 1:

Pick any two points on the arc; construct the perpendicular bisector of the line between them. Do the same for another two points on the arc; where these two lines meet is the centre of the circle.

Solution 2:

Take any three arbitrary point on arc. Now, join these points with each other. Now, Draw Perpendicular Bisectors of these lines. The intersection point of these Perpendicular bisectors would be centre of circle of which the arc is the given.. Since, Any point of perpendicular bisector of Any line segment is equidistant from its end points.

Solution 3:

Here is an alternative that works not only for finding the center of a circle from one of its arcs, but also for finding the center of an ellipse:

Draw any two non-parallel chords with endpoints on the arc. For each of these, draw a second chord parallel to the first, also with endpoints on the arc. Construct the midpoint for each chord. For each pair of parallel chords, draw the line through their midpoints. These two lines will intersect at the center of the circle (or ellipse).

This works for circles because the lines through the midpoints of parallel chords are perpendicular bisectors of those chords, hence pass through the center. It works in general because an ellipse is a linear transformation of a circle. Linear transformations don't (in general) preserve perpendicularity, which most of the other answers given here rely on, but they do preserve parallelism and length ratios (e.g. equality) of line segments along a line.

(I have to give credit for this solution to Loren Larson, who came up with it years ago when I asked him if it was possible to construct, using straightedge and compass, the center of an ellipse given only the ellipse itself.)

Solution 4:

Let $A$, $B$ and $C$ be the points on the circular arc with the center $O$ and radius $R$. Then

\begin{align} e_A&=(A_x-O_x)^2+(A_y-O_y)^2, \\ e_B&=(B_x-O_x)^2+(B_y-O_y)^2, \\ e_C&=(C_x-O_x)^2+(C_y-O_y)^2, \\ e_A&=e_B=e_C=R^2. \end{align}

Eliminating second-order terms $O_x^2$ and $O_y^2$

\begin{align} e_B-e_A&=0, \\ e_C-e_B&=0, \end{align}

we get a simple linear $2\times 2$ system in terms of coordinates $O_x$ and $O_y$ of the center:

$$ \begin{bmatrix} A_x-B_x & A_y-B_y \\ B_x-C_x & B_y-C_y \end{bmatrix} \begin{bmatrix} O_x \\ O_y\end{bmatrix} =\frac{1}{2} \begin{bmatrix} A_x^2-B_x^2+A_y^2-B_y^2 \\ B_x^2-C_x^2 +B_y^2 -C_y^2 \end{bmatrix}. $$