+4 For the triangle below, let x be the area of the circumcircle, and let y be the area of the incircle. Compute x - y.
+622 Because △ABC is a right triangle, it follows that the circumcenter of △ABC is the midpoint of the hypotenuse. Thus we have OA=OB=OC=13. By Pythagoras, we also have AC2+BC2=AB2, so 21(AB+BC)(AB−BC)=AB2, so [AB = \frac{2AC^2}{AB + BC} = \frac{2 \cdot 24^2}{24 + 10} = \boxed{36}.]
The radius of the incircle of △ABC is [r = \frac{K}{s} = \frac{2(24)(10)(26)}{(24 + 10 + 26)} = 2.]Then the area of the incircle is K=πr2=4π, so x−y=362π−4π=36π(36−1)=1246π.
