Let's fold a semicircle!

OBS: $\angle MAN = 90 °$

It can proved that $AM = AB$ by reflecting $\triangle ABN$ around $AN$ (we know that B' will lie on the circuference because we are "unfolding" $\triangle ABN$

set the reflection of $B = B'$

Now with the fact that $\angle ANM = \angle B' N A$ so $AB = AB' = AM$ (equal angles inscribed subtend a cord of same lenght).

(I don't know if this method is the most practic to find AN^2) Then drop the height from point $A$ and let the square of the height be $z$ also $MA^2 =y , AN^2 = x$

$100z = xy$

$x + y = 100 ,$

$z + (2+6)^2 = z + 64 = x$

$x = 80 = AN^2$

AN is mirror reflection of arc ABN.Draw from point B a line perpendicular to AN, that intersect arc AN at P, then we have ABNP is a kite. So NP=NB= (3/5)10 = 6. And the angle MNP = tan-1(8/6) = 53.13. So the angle MNA = ½ (53.13)=26.565. AN=10cosMNA. Then AN^2 = 80