Answer: 130

Small Hint:

Put the centers at $(-6,0)$ and $(6,0)$ and find the coordinates of $P$

Big Hint:

If $Q=P-\ell u$ and $R=P+\ell u,$ subtract the two equal-radius equations

Solution:

Put the center of the radius-$8$ circle at $O_1=(-6,0)$ and the other center at $O_2=(6,0).$ Subtracting the two circle equations gives $$P=\left(\frac76,\frac{\sqrt{455}}6\right),$$ where the positive $y$-coordinate selects the pictured intersection.

Let the common chord length be $\ell,$ and let $u$ be the unit vector from $Q$ toward $R.$ Then $Q=P-\ell u$ and $R=P+\ell u.$ Comparing $Q$ with $P$ in the first circle and $R$ with $P$ in the second gives $$\begin{aligned} u\mathbin{\cdot}(P-O_1)&=\frac{\ell}{2},\\ u\mathbin{\cdot}(P-O_2)&=-\frac{\ell}{2}. \end{aligned}$$ Adding shows that $u\perp P,$ so $$u=\frac{(\sqrt{455},-7)}{\sqrt{504}}.$$ Subtracting the two dot-product equations gives $$\ell=u\mathbin{\cdot}(O_2-O_1) =\frac{12\sqrt{455}}{\sqrt{504}}.$$ Therefore $\ell^2=\frac{144\cdot455}{504}=130.$