My geometry teacher challenged me to find a way to find the center of a given circle using only one circle and four lines. No tangents either. I can't seem to find a way to do this, all I've found so far are different ways of finding the diameter of the circle with one line left. I have always get one line left to draw, and through my own constructions,there aren't any points that I could use to draw another straight line through the diameter to find the center. This is Euclidean Geometry by the way.

A simpler version of the paragraph above is to simplify the construction of Book Three Proposition one in Euclidean Geometry so that the construction takes 4 straight lines and one circle.

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    $\begingroup$ draw two chords, find midpoints and draw perpendiculars to chords at midpoints, they will intersect at the center $\endgroup$
    – Vasili
    Feb 16, 2019 at 2:51
  • $\begingroup$ According to my teacher, it takes two circles to find the center of a line and three circles to draw a line perpendicular. And since I'm limited to one circle, this won't work. $\endgroup$
    – Pogitroy
    Feb 16, 2019 at 3:04
  • $\begingroup$ It takes two circles to draw perpendicular bisector to a segment $\endgroup$
    – Vasili
    Feb 16, 2019 at 3:49
  • $\begingroup$ Pick an arbitrary point A on given circle. Make ONE chord AB on the given circle with A and B as endpoints. Then construct circle (A; AB). Then where the given circle and (A; AB) intersect (besides B) mark it C. THEN continue getting the perpendicular bisectors of the two chords and where they intersect we have the center $\endgroup$ Feb 16, 2019 at 5:02
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    $\begingroup$ Remember, I only get one circle and four lines. Perpendicular bisectors take two circles each and since I am limited to one circle, this won't work either. $\endgroup$
    – Pogitroy
    Feb 16, 2019 at 5:13

2 Answers 2


The solution below is based on the assumption that drawing a line parallel to the other requires no auxiliary circle(s).

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Step 0: Select a random point A on the given circle.

Step 1: Draw chord AB of reasonable length.

step 2: Draw $C_1$ (centered at A, radius = AB).

Step 3: Locate C where BA produced cuts $C_1$.

step 4: The two circles will meet at D. CD, when joined, will cut the given circle at E.

Step 5: Note that $\angle CDB = \angle EDB = 90^0$ because BC is the diameter of $C_1$.

Step 6: Join BE which is then the diameter of the given circle.

Step 7: Through A, draw a line parallel to CD cutting BE at F. By intercept theorem, FB = FE.

[Note:- Drawing a line parallel to a given line can be done by translating "set squares, or straight rulers" without drawing auxiliary circles. Whether the construction of such line is considered as a 1-step construction or not depends heavily on the rule of the game.]

  • $\begingroup$ May I ask how you drew line AF? Or how you even got point F? If step 0 is drawing line AF through the center of circle ABDE parallel to line DC, then I'm afraid this wont work for me. $\endgroup$
    – Pogitroy
    Feb 20, 2019 at 3:32
  • $\begingroup$ @Pogitroy See the added steps and the comment. $\endgroup$
    – Mick
    Feb 20, 2019 at 15:41

It is actually not possible. My teacher said that after a little bit of thought, he thinks it is not possible.


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