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Circle through 3 Points
Geometry construction using a compass and straightedge
Step-by-step Instructions Printer friendly version
After doing this Your work should look like this

We start with three given points A,B,C.

Goal: Construct a circle that passes through all three.

Geometry construction with compass and straightedge or ruler or ruler
1.  (Optional*) Draw straight lines to create the line segments AB and BC. Any two pairs of the points will work.

* We draw the two lines to make it clear when we later draw their perpendicular bisectors, but it is not strictly necessary for them to actually be there to do this.

Geometry construction with compass and straightedge or ruler or ruler
2.  Find the perpendicular bisector of one of the lines. See Constructing the Perpendicular Bisector of a Line Segment. Geometry construction with compass and straightedge or ruler or ruler
3.  Repeat for the other line. Geometry construction with compass and straightedge or ruler or ruler
4.  The point where these two perpendiculars intersect is the center of the circle we desire. Geometry construction with compass and straightedge or ruler or ruler
5.  Place the compass point on the intersection of the perpendiculars and set the compass width to one of the points A,B or C. Draw a circle that will pass through all three. Geometry construction with compass and straightedge or ruler or ruler
6.  Done. The circle drawn is the only circle that will pass through all three points. Geometry construction with compass and straightedge or ruler or ruler

Explanation

This is virtually the same as constructing the circumcircle a triangle. If you draw three lines linking the given points, you will get a triangle. The circumcircle passes through all three vertices, just as here.

Proof

The image below is the final drawing above with the red items added.

  Argument Reason
1 JK is the perpendicular bisector of AB. By construction. For proof see Constructing the perpendicular bisector of a line segment
2 Circles exist whose center lies on the line JK and of which AB is a chord. (* see note below) The perpendicular bisector of a chord always passes through the circle's center.
3 LM is the perpendicular bisector of BC. By construction. For proof see Constructing the perpendicular bisector of a line segment
4 Circles exist whose center lies on the line LM and of which BC is a chord. (* see note below) The perpendicular bisector of a chord always passes through the circle's center.
5 The point O is the center of the only circle that passes through A,B,C. O is the only point that lies on both JK and LM, and so satisfies both 2 and 4 above.

  - Q.E.D

* Note
Depending where the center point lies on the bisector, there is an infinite number of circles that can satisfy this. Two of them are shown on the right. Steps 2 and 4 work together to reduce the possible number to just one.

Try it yourself
Click here for a printable worksheet containing two problems that use this construction technique. When you get to the page, use the browser print command to print as many as you wish. The printed output is not copyright.

Constructions pages on this site

Lines

Angles

Triangles

Triangle Centers

Circles, Arcs and Ellipses

Polygons

Non-Euclidean constructions