Question

The line $Ax+By+C=0$ cuts the circle $x^2+y^2+ax+by+c=0$ in $P$ and $Q$.
The line $A^{\prime }x+B^{\prime }y+C^{\prime }=0$ cuts the circle $x^2+y^2+a^{\prime }x+b^{\prime }y+c^{\prime }=0$ in $R$ and $S$.
If $P,Q,R,S$ are concyclic, then show that
$\left|\begin{array}{ccc}a-a^{\prime }& b-b^{\prime }& c-c^{\prime }\\ A& B& C\\ A^{\prime }& B^{\prime }& C^{\prime }\end{array}\right|=0$.

Answer 1

Let the given circles be denoted by $S_1=0$ and $S_2=0$ and the points $P,Q,R,S$ lie on the circle say $S_3=0.PQ$ intersects both $S_1$ and $S_3$ and $RS$ intersects both $S_2$ and $S_3$.
$\therefore PQ$ is radical axis of $S_1$ and $S_3$ and $RS$ is radical axis of $S_2$ and $S_3$
$\therefore Ax+By+C=0$ is
radical axis of $S_1$ and $S_3$ and
$A^{\prime }x+B^{\prime }y+C^{\prime }=0$ is
radical axis of $S_2$ and $S_3$.
Also radical axis of $S_1$ and $S_2$ is give by
$S_1-S_2=0$
or $(a-a^{\prime })x+(b-b^{\prime })y+(c-c^{\prime })=0$.
Again from part (a) we know that the radical axis of three circles taken in pairs are concurrent.
Hence, we have
$\left|\begin{array}{ccc}a-a^{\prime }& b-b^{\prime }& c-c^{\prime }\\ A& B& C\\ A^{\prime }& B^{\prime }& C^{\prime }\end{array}\right|=0$.