Answer: E

Small Hint:

If a chord is at signed distance $y$ from the center, its half-length satisfies $h^2+y^2=r^2$

Big Hint:

Let the signed distances of the $14$- and $10$-chords differ by $6$, and subtract their equations

Solution:

Let the signed distances from the center to the $14$- and $10$-chords be $u$ and $v,$ with $u-v=6.$ Then $$r^2-u^2=7^2,\qquad r^2-v^2=5^2.$$ Hence $v^2-u^2=24,$ so $(v-u)(v+u)=24.$ Since $v-u=-6,$ we get $u+v=-4,$ and therefore $u=1$ and $v=-5.$ The midway chord is at signed distance $-2,$ while $r^2=49+1=50.$ Its squared length is $4(50-4)=184,$ so $a=184.$ Thus the correct answer is E.