Prove that there is only one circle determined by three noncollinear points

O lies on the perpendicular bisector of AB. ... Then, O' will lie on the perpendicular bisector PQ and RS. We know that, two lines cannot intersect at more than one point, So O' must coincide with O. Hence, there is one and only one circle passing through three non collinear points

Step-by-step explanation:

both equations and proportions are better.