The line $x+2y+a=0$ intersects the circle $x^2+y^2-4=0$ at two distinct points $A$ and $B$. Another line $12x-6y-41=0$ intersects the circle $x^2+y^2-4x-2y+1=0$ at two distinct points $C$ and $D$. If the four points $A,B,C$, and $D$ are concyclic then the value of $a$ is

The given circles are
$C_1:x^2+y^2-4=0L_1:x+2y+a=0$
$C_2:x^2+y^2-4x-2y+1=0L_2:12x-6y-41=0$


If $A,B,C$, and $D$ are concyclic, then
$PA.PB=PC.PD$ or
$P{T}^2=P{T}^{\prime }{}^2$
$\therefore$ Point $P$ will lie on the radical axis of both the circles $C_1$ and $C_2$,
Hence the equation of radical axis is $4x+2y-5=0$
Now, the lines
$4x+2y-5=0$
$x+2y+a=0$ and
$12x-6y-41=0$ are concurrent,Therefore,
$\left|\begin{array}{ccc}4& 2& -5\\ 1& 2& a\\ 12& -6& -41\end{array}\right|=0$
$\therefore a=2$