Unique circle passing through three non-collinear point

A unique circle can be drawn through any three non-collinear points in a plane.
circle passing through three points

This theorem confirms the existence and uniqueness of a circle that intersects three non-collinear points.

A Practical Example

Start by plotting three non-collinear points.

three non-collinear points

Connect points A and B to form segment AB.

Then, find the midpoint, M, of segment AB and construct the perpendicular bisector of the segment through M.

perpendicular bisector of segment AB

Next, connect points B and C to form segment BC.

Similarly, find the midpoint of segment BC and draw its perpendicular bisector.

perpendicular bisector of segment BC

The bisectors of segments AB and BC intersect at point O, equidistant from A, B, and C.

point O

Explanation: Perpendicular bisectors of a segment include all points equidistant from the segment's endpoints. Thus, point O, lying at the intersection of the bisectors for AB and BC, is equidistant from all three points.

Now, draw a circle centered at O with radius equal to OA (OB or OC would be equivalent).

circle passing through three non-collinear points

The result is a circle that perfectly passes through points A, B, and C, demonstrating the theorem.

Not Just Existence, but Uniqueness: This theorem doesn’t merely establish that such a circle exists; it also proves that no other distinct circle can encompass A, B, and C. The intersecting bisectors uniquely define point O, precluding any other point from being equidistant from A, B, and C. More details follow in the demonstration.

The Proof

The theorem asserts the existence and uniqueness of a circle that intersects three non-aligned points A, B, and C in the plane.

Existence

Point O resides on the perpendicular bisectors of both segments AB and BC.

perpendicular bisector of segment BC

As a geometric locus, the perpendicular bisector contains all points equidistant from the segment's endpoints, confirming that segments OA and OB are congruent, as are OB and OC.

$$ \overline{OA} \cong \overline{OB} $$

$$ \overline{OB} \cong \overline{OC} $$

By transitivity, if OA = OB and OB = OC, then OA = OC.

$$ \overline{OA} \cong \overline{OC} $$

Therefore, point O maintains equal distances from A, B, and C.

point O

This implies that A, B, and C all belong to the circle centered at point O.

Uniqueness

Since A, B, and C are not aligned, segments AB and BC are not parallel and thus intersect.

The lines through segments AB and BC are intersecting lines, intersecting at a unique point, point B.

intersecting lines

The perpendiculars to intersecting lines also intersect, confirming that the perpendicular bisectors of segments AB and BC intersect at only one point.

circle passing through three points

Thus, point O is uniquely equidistant from A, B, and C. There can be no others.

Conclusion

In conclusion, a single circle uniquely passes through the three points A, B, and C, as demonstrated by this theorem.

And so on.

 
 

Please feel free to point out any errors or typos, or share suggestions to improve these notes. English isn't my first language, so if you notice any mistakes, let me know, and I'll be sure to fix them.

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