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.
 A: 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.
A: 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:
$$
\begin{bmatrix}
A_x-B_x & A_y-B_y \\
B_x-C_x & B_y-C_y 
\end{bmatrix}
\begin{bmatrix} O_x \\ O_y\end{bmatrix}
=\frac{1}{2}
\begin{bmatrix} 
A_x^2-B_x^2+A_y^2-B_y^2 \\
B_x^2-C_x^2 +B_y^2 -C_y^2
\end{bmatrix}.
$$
A: 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. 
A: This can be done by taking 3 points on the arc and drawing the perpendicular bisectors of the line joining these points,the point of intersection is the center
Let me show it with an example:
Take three points A,B and C taken in order on the arc.
Draw the perpendicular bisector of AB and BC.
The point at which the perpendicular bisectors intersect is the center
A: 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.)
A: You can use the "Set of Thales" or called "Thales Theorem." There is a good drawing of it here:
Set of Thales.
A: If you have an arc, you can pick three points on the arc and find what is called the circumscribed circle of the triangle.  This wikipedia article covers the relevant formulas for the centre and radius of the circle.
A: Let pq be the arc,take the point 'R'(take the point R on the arc a bit far from P&Q) on the arc and join PR,PQ. Now draw perpendicular bisectors of PR & PQ and extend these till they intersect. Now the point of intersection of these bisectots will be the CENTER....
A: If the given arc is large enough (in the sense of longer than the semi-circle), a large enough A4 sheet can do the required job. See Fig.1.

If the original arc is not long enough, it can be extended continuously by the method described in Fig. 2 until the above requirement is met.
Note also that the line of symmetry can be created by folding up the A4 paper.
A: center, 1. Geometry. the middle point, as the point within a circle or sphere, usually distant from all  points of the circumference or surface. 2. a point, pivot, axis, etc., around which anything rotates or revolves. 7. the core or middle of anything. 14. Math. the mean position of a figure or system. 15. Mach. a tapered rod, mounted in the headstock spindle (live center) or the tail stock(dead center) of a lathe, upon which a work is placed., central or normal position.      polar coordinate system. the frame of reference is a horizontal halfline or ray and its starting point. The ray is called the polar axis and its starting point is called the pole. A point is located by giving its distant r from the pole and the angle 0 from the polar axis to the ray from the pole through the point. The directed distance r is called the radius vector and the angle 0 is known as the vectorial angle or amplitude. The vectorial angle may be positive, zero, or negative. Distances measured from the pole  along the radius vector are positive and along the extension of the radius vector through the pole are negative. If a value r of the radius vector and a value 0 of the vectorial angle are given, exactly one point is determined by the ordered number pair (r,0). The perpendicular to the polar axis through the pole is called the normal axis but it is not used in locating points in polar coordinates.
