Which point of the plane is equidistant from the three points A(2, 3), B(8, 3) and C(8, 11)?

AB is horizontal and BC is vertical, so the angle at B is a right angle and the required point — the circumcentre — is the midpoint of the hypotenuse AC, namely ((2 + 8)/2, (3 + 11)/2) = (5, 7). Confirm all three distances: to A, √$(3^2+4^2)=5$; to B, √$(3^2+4^2)=5$; to C, √$(3^2+4^2)=5$. The trap (6, 17/3) is the CENTROID, the average of the three vertices, which is equidistant from nothing in particular; (8, 7) is the midpoint of BC and (5, 3) the midpoint of AB — each serves only two of the three points.