Let the two circles have centers $O_1$ and $O_2$, with radii $r_1=5,\\ r_2=12.$

The circles intersect at a point $P$, and the angle of intersection is $90^\circ$.

The angle of intersection of two circles is the angle between their tangents at the point of intersection. This is equal to the angle between the radii drawn to that point.

Hence, $\angle O_1PO_2=90^\circ.$

In triangle $O_1PO_2$, we have: $O_1P=5, \quad O_2P=12, \quad \angle O_1PO_2=90^\circ.$

So by Pythagoras, $O_1O_2=\sqrt{5^2+12^2}=\sqrt{25+144}=13.$

Let $AB$ be the common chord of the two circles, and let $M$ be its midpoint.

The common chord is perpendicular to the line of centers $O_1O_2$. So $M$ lies on $O_1O_2$.

Let $O_1M=x.$ Then $O_2M=13-x.$

Since $M$ is the midpoint of the common chord, half the chord length is $AM=BM=\frac{AB}{2}.$

Using circle 1: $x^2+\left(\frac{AB}{2}\right)^2=5^2=25. \\ (1)$

Using circle 2: $(13-x)^2+\left(\frac{AB}{2}\right)^2=12^2=144. \\ (2)$

Subtract (1) from (2): $(13-x)^2-x^2=144-25=119.$

Expanding, $169-26x=119$ $26x=50$ $x=\frac{25}{13}.$

Substitute into (1): $\left(\frac{25}{13}\right)^2+\left(\frac{AB}{2}\right)^2=25$ $\frac{625}{169}+\left(\frac{AB}{2}\right)^2=25$

So the length of the common chord is $\boxed{\frac{120}{13}}.$

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