Longest Chord in a circle?

Question

A rectangle with side lengths $AB = 3$ and $BC = 11$ is inscribed within a circle, as shown. There are two chords $\overline{AS}$ of the circle which are bisected by segment $\overline{BC}$. Find the length of the longer such chord.

Answer 1

Let $BM = x$. Then $AM^2 = 9 + x^2$ (Pythagoras) and if $M$ bisects $AS$ then also $AM^2 = x(11 - x)$ (intersecting chords). Hence $(x - 1)(2x - 9) = 0$, giving $AS = 2\sqrt10$ or $3\sqrt13$.

Answer 2

we can calculate the length of $ \overline{AM} $ by using the Pytagoras Theorm ( $\overline{AB}^2+\overline{BM}^2=\overline{AM}^2$ )
Lastly, calculate $\overline{MS}$ by $\overline{MS} \ \overline{AM}=\overline{BM}\overline{MC}$
then you know which is bigger and how long it is.