Here is a circumscribed circle to the given right triangle.

If you recall your geometry about inscribed angle, it is half the measure of its intercepted arc.

# m∠ACB = ½ m (arc AB)

90° = ½ m (arc AB)

m(arc AB) = 180°

It only shows that arc AB is a semi-circle and the diameter is AB which is also the hypotenuse of the right triangle. To get the radius, simply get the half of the measure of the diameter AB.

(AB)^{2} = 4^{2} +7^{2}

(AB)^{2} =16 + 49

(AB)^{2} = 65

AB = (65)^{1/2}

½ AB = ½ (65)^{1/2} ≈ **4.03 cm**