The line Ax+By+C=0 cuts the circle x²+y²+g x+f y+c=0 at P and Q. The line A' x+B' y+C'=0 cuts the circle ${\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{g}}^{\text{'}}\mathrm{x}+{\mathrm{f}}^{\text{'}}\mathrm{y}+{\mathrm{c}}^{\text{'}}=0$ at R and S. If P, Q, R and S are concyclic, then the value of the determinant $\left|\begin{array}{ccc}\mathrm{g}-\mathrm{g}\text{'}& \mathrm{f}-\mathrm{f}\text{'}& \mathrm{c}-\mathrm{c}\text{'}\\ \mathrm{A}& \mathrm{B}& \mathrm{C}\\ \mathrm{A}\text{'}& \mathrm{B}\text{'}& \mathrm{C}\text{'}\end{array}\right|$

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 $S_3$​.
∴PQ is radical axis of ​ $S_1$ and  $S_3$​ and RS is radical axis of $S_2$​ and $S_3$
∴Ax+By+C=0 is radical axis of ​$S_1$ and ​$S_3$ and
A′x+B′y+C′=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
$\Rightarrow$(g-g')x+(f-f')y+(c−c′)=0.
we know that the radical axis of three circles taken in pairs are concurrent.

$\left|\begin{array}{ccc}g-g\text{'}& f-f\text{'}& c-c\text{'}\\ A& B& C\\ A\text{'}& B\text{'}& C\text{'}\end{array}\right|=0$