29
$\begingroup$

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.

$\endgroup$
3
  • 5
    $\begingroup$ If you know it is an actual arc of a circle and if you know several points on it then you can write several equations of the form $\;(x_i-h)^2+(y_i-k)^2=R^2\;$, with $\;(x_i,y_i)\;$ points on the arc, until you find the center $\;(h,k)\;$ . $\endgroup$
    – Timbuc
    Jan 21 '15 at 9:51
  • 3
    $\begingroup$ can you draw the normals at two points? see where they intersect. will that work? $\endgroup$
    – abel
    Jan 21 '15 at 9:52
  • 4
    $\begingroup$ A question with 22 upvotes and 8 answers "has not received enough attention" ? $\endgroup$ Jan 26 '15 at 17:31

10 Answers 10

87
$\begingroup$

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.

$\endgroup$
4
  • 35
    $\begingroup$ +1. Hence, even if we have only three points of the circle, we can determine the circle uniquely. $\endgroup$
    – orangeskid
    Jan 21 '15 at 9:54
  • $\begingroup$ Ah, this is way easier than my original idea. +1 $\endgroup$
    – Timbuc
    Jan 21 '15 at 9:55
  • $\begingroup$ You can add the point with some clarification/reference raised by @orangeskid as essentially this is what you are using: three distinct points in a plane determines a circle uniquely. Really nice answer. +1 $\endgroup$
    – Krish
    Jan 21 '15 at 10:13
  • 9
    $\begingroup$ @user73985: +1 . Make sure the chords are not parallel, for this, reusing one end point will do. $\endgroup$
    – orangeskid
    Jan 21 '15 at 10:19
24
$\begingroup$

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.enter image description here

$\endgroup$
2
  • 5
    $\begingroup$ :+1, great. Also, you don't have to worry about end points, just don't pick any parallel chords. $\endgroup$
    – orangeskid
    Jan 21 '15 at 10:17
  • 9
    $\begingroup$ The picture definitely helps understand why this works. I think a thinner pen stroke would help the clarity of the picture. $\endgroup$
    – Brian J
    Jan 21 '15 at 21:33
19
+100
$\begingroup$

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.)

$\endgroup$
1
  • 4
    $\begingroup$ Upvote. This is way more interesting and useful than the perpendicular bisector trick I learned in 8th grade. $\endgroup$
    – imallett
    Jan 22 '15 at 5:22
6
$\begingroup$

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:

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

$\endgroup$
2
  • 1
    $\begingroup$ +1 for giving an algebraic method (since others have covered the geometric construction just fine). $\endgroup$ Jan 22 '15 at 5:04
  • $\begingroup$ Thank you @g.kov for this easy to follow answer. I have created a PowerShell script that takes three points and works out the radius and center of the circle: gist.github.com/tahir-hassan/0b1720be42bdcc9c6a06fdd864f3e7f7 $\endgroup$ Jul 18 at 17:06
2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

You can use the "Set of Thales" or called "Thales Theorem." There is a good drawing of it here: Set of Thales.

$\endgroup$
1
  • 1
    $\begingroup$ But what if the arc is shorter than a half of the full circle? I.e., what if for any two points $A,B$ on the arc, the line $AC (AC\perp AB)$ does not meet the arc at any point $M$ ($M\neq A$)? $\endgroup$
    – user26486
    Jan 21 '15 at 22:19
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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....

$\endgroup$
0
$\begingroup$

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.

enter image description here

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.