The longer chord is farther from the center of the circle than the shorter chord.
This can be proven using the following theorem:
Theorem: If two chords of a circle are congruent, then the longer chord is farther from the center of the circle than the shorter chord.
Proof:
Let $AB$ and $CD$ be two congruent chords of a circle with center $O$.
Since $AB$ and $CD$ are congruent, then $|AB| = |CD|$.
Let $d_1$ be the distance from $O$ to $AB$ and $d_2$ be the distance from $O$ to $CD$.
Since $O$ is the center of the circle, then $d_1 = d_2$.
Now, let $E$ be the midpoint of $AB$ and $F$ be the midpoint of $CD$.
Since $E$ is the midpoint of $AB$, then $|AE| = |EB| = \frac{1}{2}|AB|$.
Since $F$ is the midpoint of $CD$, then $|CF| = |FD| = \frac{1}{2}|CD|$.
Since $|AB| = |CD|$ and $E$ and $F$ are the midpoints of $AB$ and $CD$, respectively, then $|AE| = |EB| = |CF| = |FD|$.
Since $|AE| = |CF|$ and $d_1 = d_2$, then $|AO| = |OC|$.
Therefore, $O$ is equidistant from $AB$ and $CD$.
Since $O$ is equidistant from $AB$ and $CD$, then the longer chord $CD$ is farther from the center of the circle than the shorter chord $AB$.