P(-2, 0) and R(6, 0) are the ends of a diameter of a circle. Does the point Q(2, 3) lie on that circle?

The centre is the midpoint of the diameter, $\left(\frac{-2+6}{2},\frac{0+0}{2}\right)=(2,0)$, and the radius is half of PR = |$6-(-2)$| = 8, so $r=4$. Now $CQ=\sqrt{(2-2{)}^2+(3-0{)}^2}=3$. Since 3 < 4, $Q$ lies inside the circle, not on it. Comparing $Q$ only with P or R (for example PQ = 5) does not answer the question: what matters is the distance from the centre.