A circle centered at $O$ has radius $10$ ft. A point is outside the circle if its distance from $O$ exceeds $10$.

Use the Pythagorean theorem to find each point's distance from $O$ (each point is the hypotenuse vertex of a right triangle with legs as labeled):

• $A$: $\sqrt{4^2 + 9^2} = \sqrt{16 + 81} = \sqrt{97} \approx 9.85$ (inside)

• $B$: $\sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$ (on the circle)

• $C$: $\sqrt{7^2 + 7^2} = \sqrt{49 + 49} = \sqrt{98} \approx 9.90$ (inside)

• $D$: $\sqrt{6^2 + 10^2} = \sqrt{36 + 100} = \sqrt{136} \approx 11.66$ (outside)

• $E$: $\sqrt{1^2 + 10^2} = \sqrt{1 + 100} = \sqrt{101} \approx 10.05$ (outside)

Point $E$ with legs $1$ and $10$ has distance $\sqrt{101} > 10$, so it is outside the circle.