Concyclicity of $4$ points using algebraic geometry 
Consider the line $L_1:ax+4y-1=0$ and a circle $S:x^2+y^2-10x+2y+10=0$. The line intersects the circle at $2$ distinct points $A$ and $B$. Another line $5x-12y-67=0$ intersects the circle $x^2+y^2+6x+14y-28=0$ at $2$ distinct points $C$ and $D$. Find the value of $a$ for which the $4$ points $A,B,C,D$ are concyclic. Also, find the equation of the circle passing through these $4$ points.

How do I use the concyclic condition? If I make equations using opposite angles sum $180°$, the equations are pretty nasty. Some hints, please. Thanks.

HINT:
The equation of the circle passing through the intersection of $L_1, S$
$$x^2+y^2-10x+2y+10+A(ax+4y-1)=0\  \ \ \ (1)$$
Similarly for $5x-12y-67=0, x^2+y^2+6x+14y-28=0$
$$x^2+y^2+6x+14y-28+B(5x-12y-67)=0\  \ \ \ (2)$$
$A,B,C,D$ will be concyclic if $(1)$ & $(2)$ become identical.  
Can you take it from here?
