jmc

Rewrite the given equations as $(x+5{)}^2+(y-12{)}^2=256$ and $(x-5{)}^2+(y-12{)}^2=16$. Let $w_3$ have center $(x,y)$ and radius $r$. Now, if two circles with radii $r_1$ and $r_2$ are externally tangent, then the distance between their centers is $r_1+r_2$, and if they are internally tangent, it is $|r_1-r_2|$. So we have $$\begin{array}{rl}r+4& =\sqrt{(x-5{)}^2+(y-12{)}^2}\\ 16-r& =\sqrt{(x+5{)}^2+(y-12{)}^2}\end{array}$$ Solving for $r$ in both equations and setting them equal, then simplifying, yields $$\begin{array}{rl}20-\sqrt{(x+5{)}^2+(y-12{)}^2}& =\sqrt{(x-5{)}^2+(y-12{)}^2}\\ 20+x& =2\sqrt{(x+5{)}^2+(y-12{)}^2}\end{array}$$ Squaring again and canceling yields $1=\frac{x^2}{100}+\frac{(y-12{)}^2}{75}.$ So the locus of points that can be the center of the circle with the desired properties is an ellipse. Since the center lies on the line $y=ax$, we substitute for $y$ and expand:$$1=\frac{x^2}{100}+\frac{(ax-12{)}^2}{75}⟹(3+4a^2)x^2-96ax+276=0.$$ We want the value of $a$ that makes the line $y=ax$ tangent to the ellipse, which will mean that for that choice of $a$ there is only one solution to the most recent equation. But a quadratic has one solution iff its discriminant is $0$, so $(-96a{)}^2-4(3+4a^2)(276)=0$. Solving yields $a^2=\frac{69}{100}$, so the answer is $\boxed{169}$.