A circle has centre O(1, 2) and radius 5 units. Where does the point P(4, 7) lie?

$OP=\sqrt{(4-1{)}^2+(7-2{)}^2}=\sqrt{9+25}=\sqrt{34}\approx 5.83$. Compare this with the radius 5: $OP$ is greater, so $P$ lies outside the circle. A quicker check avoids surds altogether: $O{P}^2=34$ and $r^2=25$, and 34 > 25. Reading 9 + 25 as roughly 5 and stopping there is what leads to the "on the circle" answer.