How many distinct circles can be drawn through three given non-collinear points in a plane?

Introduction / Context:
This is a fundamental concept question in Euclidean geometry about circles and points. It focuses on the uniqueness of a circle passing through three non-collinear points. Understanding this fact is essential for many constructions and proofs related to circumcircles of triangles and circle geometry in general.

Given Data / Assumptions:

• We are given three points in a plane.
• The three points are non-collinear, meaning they do not all lie on the same straight line.
• We want to know how many different circles can pass through all three points.
• Standard Euclidean geometry is assumed.

Concept / Approach:
In Euclidean geometry, any three non-collinear points uniquely determine a circle, called the circumcircle of the triangle formed by those points. This happens because there is exactly one circle whose centre is equidistant from all three points. That centre is the point where the perpendicular bisectors of the sides of the triangle intersect, which is a unique point for non-collinear points. Therefore, only one circle can pass through all three such points.

Step-by-Step Solution:
Step 1: Think of the three non-collinear points as the vertices of a triangle, say A, B and C. Step 2: For a circle to pass through A, B and C, its centre must be equidistant from all three points. Step 3: The set of points equidistant from A and B forms the perpendicular bisector of segment AB. Step 4: Similarly, the set of points equidistant from B and C forms the perpendicular bisector of segment BC. Step 5: For a point to be equidistant from A, B and C, it must lie on both perpendicular bisectors AB and BC. Step 6: In a plane, two non-parallel lines intersect at exactly one point. The perpendicular bisectors of AB and BC are not parallel when the points are non-collinear. Step 7: Therefore, the perpendicular bisectors intersect at exactly one point, which is the unique centre of the circumcircle. Step 8: With this unique centre and the common distance to A, B and C as radius, exactly one circle passes through the three non-collinear points.

Verification / Alternative check:
If you attempted to draw a second circle through the same three non-collinear points, its centre would also have to be equidistant from all three points. That would imply a second intersection point of the same perpendicular bisectors, which is impossible because two non-parallel lines intersect at only one point. Hence a second distinct circle cannot exist, reinforcing that there is exactly one.

Why Other Options Are Wrong:
Two, three or more than three: These options contradict the uniqueness of the intersection of the perpendicular bisectors and the requirement that a circle's centre must be equidistant from all three points.
None: This would be true only if the points were collinear. But the question explicitly states that the points are non-collinear, so a circle does exist.

Common Pitfalls:
A misunderstanding often arises when students confuse the case of collinear and non-collinear points. For three collinear points, no circle can pass through all of them, but for non-collinear points, exactly one circle does. Another pitfall is overlooking the role of perpendicular bisectors and the uniqueness of their intersection. Remembering the circumcircle of a triangle helps solidify this concept.

Final Answer:
The number of circles that can be drawn is exactly one.