Given three points, it is possible to draw a circle that passes through all three*. This page shows how to construct (draw) a circle through 3 given points with compass and straightedge or ruler. It works by joining two pairs of points to create two chords. The perpendicular bisectors of a chords always passes through the center of the circle. By this method we find the center and can then draw the circle.
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.
If the three points are collinear (all lying on a straight line) then the circle passing through all three will have a radius of infinity. So there is no practical circle that can pass through three collinear points. If you tried the construction, you would find that the two radii (JO, LO) would be parallel and so never meet at a center. In the language of mathematics, we might say they do intersect, but at infinity.
The above animation is available as a printable step-by-step instruction sheet, which can be used for making handouts or when a computer is not available.
The image below is the final drawing above with the red items added.
|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.|
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 below. Steps 2 and 4 work together to reduce the possible number to just one.