## The center of a circle

Hi everybody,

I ve got a question : is it possible to identify the center of a given cirle with only a compass?

 Quote by pixel01 I ve got a question : is it possible to identify the center of a given cirle with only a compass?
No you need a ruler as well.

Pick a point A on the circle. Draw a circle with center A and diameter less than that of the given circle. Mark off the two points B and C where this new circle intersects the given one. Join B and C with the line BC.

Bisect BC. The midpoint is D. Draw a straight line through A and D and extend it until it meets the given circle again at E. AE is a diameter.

Repeat this process for a second point on the given circle. Where the two diameters meet is the center.

 Thank you for answering. But the problem is without a ruler !!! It seems impossible. I have tried many times but failed.

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## The center of a circle

It works without a ruler, though only for a circle that's on a sheet of paper.
Instead of using a ruler fold the paper to make a line visible.

 Quote by Edgardo It works without a ruler, though only for a circle that's on a sheet of paper. Instead of using a ruler fold the paper to make a line visible.
Well, it seems like a trick ! Let's say the circle is on a table. May be we need to prove it impossible to do so.

 It's great !. Thank you Jeroen. Anyway, how can we prove that friends?
 The method works by drawing a circle, picking two points on it, then draw a series of circles based on where the existing circles. The final answer is the intersection of the last two circles. All you need to do is write a big formula containing all these circles and resulting in the coordinates of the final intersection and then prove that "the_big_function(center_of_circle, radius_of_circle, point_1_on_circle, point_2_on_circle) = center_of_circle" is true for any input. This would be a start: Code: point float x float y circle point center float radius point intersect (circle ca, circle cb, point pn) returns intersection between ca and cb that is not pn float distance (point pa, point pb) returns distance between points pa and pb point pointoncircle (circle c, float angle) returns a point on the circle going the given angle clockwise from the top input: circle c0 c0.center = variable c0.radius = variable float a1 a1 = variable float a2 a2 = variable output: point pf process: point p1 p1 = pointoncircle(c0,a1) point p2 p2 = pointoncircle(c0,a2) circle c1 c1.center = p1 c1.radius = distance(p1,p2) circle c2 c2.center = p2 c2.radius = distance(p1,p2) circle c3 c3.center = intersect(c0,c1,p2) c3.radius = distance(p1,p3) circle c4 c4.center = intersect(c2,c3,p1) c4.radius = distance(intersect(c2,c3,p1),p1) circle c5 c5.center = intersect(c4,c3,p1) c5.radius = distance(intersect(c4,c3,p1),p1) circle c6 c6.center = intersect(c2,c4,p1) c6.radius = distance(intersect(c2,c4,,p1),p1) pf = intersect(c5,c6,p1)
 Recognitions: Gold Member Science Advisor Staff Emeritus I may be wrong but I seem to remember a theorem that any construction that could be done with compasses and straight edge could be done with compasses alone. Of course "drawing a line" has to be interpreted as constructing two points on the line. Check: http://thesaurus.maths.org/mmkb/entr...ryById&id=4066