Solution

First consider just two of the three points, say P and Q, and the set of points in the plane that are the same distance from them. Clearly the midpoint of line segment PQ is such a point. Are there others? You may recall from geometry that the points on the line that bisects PQ and is perpendicular to PQ are all the points that are equidistant from P and Q. Similarly, the points in the plane that lie on the perpendicular bisector of line segment PR are all the points that are equidistant from points P and R.

Because R does not lie on line PQ, line segments PQ and PR do not lie on the same line, and so their respective perpendicular bisectors are not parallel.
Therefore, you can conclude that the two perpendicular bisectors intersect at a point. The point of intersection is on both perpendicular bisectors, so it is equidistant from P and Q as well as from P and R. Therefore, the point of intersection is equidistant from all three points. Are there any other points that are equidistant from P, Q, and R ? If there were, they would be on both perpendicular bisectors, but in fact only one point lies on both lines.

The answer is \(B\)
_________________

Get the 2 FREE GREPrepclub Tests

If R is not on line PQ, then the three points form the vertices of a triangle. Let’s call it triangle PQR. Since there is only one point that is equidistant (the same distance) from the three vertices of the triangle (and the point is actually called the circumcenter), choice B is the correct answer.

Answer: B
_________________

yes. For sure.

it is an official question from the Official Guide.

Regards
_________________

Get the 2 FREE GREPrepclub Tests

Another useful way to think about this problem is to realize that three points determine a circle. We can draw a circle such that points P, Q, and R all lie on the circumference, with the circle's center being the same distance (radius) from all three points. Since the points are not colinear only one such circle is possible.