The line $$x + 2y + a=0$$ intersects the circle $$x^2+ y^2= 4$$ at two distinct points $A$ and $B$ another line $$12x - 6y - 41 =0$$ intersect the circle $$x^2 + y^2- 4x - 2y + 1=0$$ at two distinct points $C$ and $D$. Find the value of $a$ for which the quadrilateral $ABCD$ is a cyclic quadrilateral.
My Work
I tried doing this question the general way i.e. the Ptolemy's theorem but the points obtained on solving the lines with the circles consisted of so much of the irrational terms that I might have messed up the equations nearly seven times until now. Can someone provide some help over how to approach such question.
Note : One interesting thing I noted in this question is that the lines given to us are perpendicular to each other. Is this information helpful to solve this problem or is it a mere coincidence.
Edit 1: An idea just struck me. I think we might use the concept of family of circles. One member of the first family and the second family must be same for the points to be concyclic. But I am not able to continue with it. If anyone finds a solution in this method please share.
Let $P$ be point of intersection of the two lines and $E$ be one of the two points of intersection of the two circles.
Suppose that $PE$ meets the circle $ABE$ at $E$ and $F_1$. By the Power of a Point Theorem, $PA\cdot PB=PE\cdot PF_1$.
Suppose that $PE$ meets the circle $CDE$ at $E$ and $F_2$. By the Power of a Point Theorem, $PC\cdot PD=PE\cdot PF_2$.
If $A$, $B$, $C$ and $D$ are concyclic, $PA\cdot PB=PC\cdot PD$. This implies that $PE\cdot PF_1=PE\cdot PF_2$. So $F_1$ and $F_2$ are the same point, which is another point of intersection of the two circles.
So $P$ lies on the common chord ($4x+2y-5=0$) of the circles.
So, $x+2y+a=0$ is a member of the family of lines $12x-6y-41+k(4x+2y-5)=0$
\begin{align*} 12+4k=\frac{-6+2k}{2}=\frac{-41-5k}{a} \end{align*}
$k=-5$ and $a=2$.