(b) A and B have coordinates $(-12,8),(-0.8,14.4)$.

(a)
$$
\begin{array}{l}
(x+8)^2+(y-14)^2=52 \\
(x+8)^2+(m x-14)^2=52 \\
x^2+16 x+64+m^2 x^2-28 m x+196=52 \\
\left(1+m^2\right) x^2+(16-28 m) x+208=0 \\
\Rightarrow(16-28 m)^2-4\left(1+m^2\right)(208)=0 \\
256-896 m+784 m^2-832-832 m^2=0 \\
-48 m^2-896 m-576=0 \\
3 m^2+56 m+36=0
\end{array}
$$

(b)$3 m^2+56 m+36=0 \implies m=-\frac{2}{3}$ or $m=-18$

$m=-\frac{2}{3}:$ $\left(m^2+1\right) x^2+(16-28 m) x+208=0 \implies \frac{13}{9} x^2+\frac{104}{3} x+208=0$

$\implies x=-12, \quad y= -\frac{2}{3} (-12) = 8$

$m=-18: \left(m^2+1\right) x^2+(16-28 m) x+208=0 \implies 325 x^2+520 x+208=0$

$\implies x=-\frac{4}{5} \quad y= -\frac{4}{5} (-12) = \frac{72}{5}$

$\therefore$ A and B have coordinates $(-12,8),(-0.8,14.4)$.