Question

The line $2x+3y+4=0$ cut the circle $x^2+y^2+ax+by+c=0$ at $P$ and $Q$. The line $x-3y+2=0$ cut the circle $x^2+y^2+a^{\prime }x+b^{\prime }y+c^{\prime }$ at $R$ and $S$. If $P,Q,R$ and $S$ are concyclic and value of $\left|\begin{array}{ccc}a-a^{\prime }& b-b^{\prime }& c-c^{\prime }\\ 2& 3& 4\\ 1& -3& 2\end{array}\right|=k\left(abc\right)\left(a^{\prime }{b}^{\prime }{c}^{\prime }\right)$, then $k=$

Answer 1

$P$ and $Q$ are the points of the intersection of the $l_1:2x+3y+4=0$ and $C_1:x^2+y^2+ax+by+c=0$
and $R$ and $S$ are the points of the intersection of the $l_2:x-3y+2=0$ and $C_2:x^2+y^2+a^{\prime }x+b^{\prime }y+c^{\prime }=0$
$\because P,Q,R,S$ are concyclic
$\therefore$ the radical axis of the circles $C_1=0$ and $C_2=0$ is $l_3:C_1-C_2=0$
$l_3:(a-a^{\prime })x+(b-b^{\prime })y+c-c^{\prime }=0$
Let $C_3$ be the circle passing through $P,Q,R,S$
Then $l_1$ will be the radical axis of the $C_1=0$ and $C_3=0$ and Then $l_2$ will be the radical axis of the $C_2=0$ and $C_3=0$
$\because l_1,l_2$ and $l_3$ are the radical axes of the circles $C_1=0,C_2=0$ and $C_3=0$ taken in pairs
this is possible if all the three lines are parallel or concurrent

$\therefore \left|\begin{array}{ccc}a-a^{\prime }& b-b^{\prime }& c-c^{\prime }\\ 2& 3& 4\\ 1& -3& 2\end{array}\right|=0$