, 2016

Since $M$ is the midpoint of chord $AB$, then $AM=\frac12(AB)=5$.

Also, since $M$ is the midpoint of chord $AB$, then $OM$ is perpendicular to $AB$.
Using the Pythagorean Theorem in $\triangle OMA$, we get $OM^2=OA^2-AM^2$or $OM^2=13^2-5^2=169-25=144$, and so $OM=\sqrt{144}=12$ (since $OM>0$).
Let the circle have centre $O$ and chord $PQ$, as shown.

Since the radius is 25, then $OQ=25$.

The perpendicular distance from $O$ to the chord is given by $OR$, and so $OR=7$.
In $\triangle ORQ$, the Pythagorean Theorem gives $RQ^2=OQ^2-OR^2$ or $RQ^2=25^2-7^2=625-49=576$, and so $RQ=\sqrt{576}=24$ (since $RQ>0$).

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Since $OR$ is perpendicular to the chord $PQ$, then $R$ is the midpoint of $PQ$, and so $PQ=2(RQ)=2(24)=48$.
Therefore, the length of the chord is 48.
Join $O$ to $S$ and $O$ to $U$, as shown.

The radius of the circle is 65, and so $OS=OU=65$.

Since $OM$ is perpendicular to chord $ST$, then $M$ is the midpoint of the chord and so $MS=\frac12(ST)=\frac12(112)=56$.

In $\triangle OMS$, the Pythagorean Theorem gives $OM^2=OS^2-MS^2$ or $OM^2=65^2-56^2=4225-3136=1089$, and so $OM=\sqrt{1089}=33$ (since $OM>0$).

Since $MN=OM+ON=72$, then $ON=72-OM=72-33=39$.
In $\triangle ONU$, the Pythagorean Theorem gives $NU^2=OU^2-ON^2$

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or $NU^2=65^2-39^2=4225-1521=2704$, and so $NU=\sqrt{2704}=52$ (since $NU>0$). Finally, since $ON$ is perpendicular to chord $UV$, then $N$ is the midpoint of the chord and so $UV=2(NU)=2(52)=104$.

Therefore, the length of the chord $UV$ is 104.