The line `A x+B y+C=0` cuts the circle `x^2+y^2+a x+b y+c=0` at `Pa n dQ` . The line `A^(prime)x+B^(prime)x+C^(prime)=0` cuts the circle `x^2+y^2+a^(prime)x+b^(prime)y+c^(prime)=0` at `Ra n dSdot` If `P ,Q ,R ,` and `S` are concyclic, then show that `|a-a ' b-b ' c-c ' A B C A ' B ' C '|=0`

Explanation

P and Q are the points of intersection of the line
`L_(1)-=Ax+By+C=0` (1)
and the circle
`S_(1)-= x^(2)+y^(2)+ax+by+c=0` (2)
R and S are the ponits of intersection of the line
`L_(2)-=A'x+B'y+C'=0` (3)
and the circle
`S_(2) -= x^(2)+y^(2)+a'x+b'y+c'=0` (4)

The radical axis of the circles `S_(1)=` and `S_(2)=` is
`S_(1)- S_(2)=0`
i.e., `L_(3)-= (a-a')x+(b-b')y+(c-c')=0` (5)
If P,Q,R, and S are concyclic and `S_(3)=` is the equation of this circle thorugh P,Q,R, and S, line (1) is the radical axis of the circle `S_(1)=0` and `S_(3)=0` and line (2) is the radical axis of the circles `S_(2)` and `S_(3)=`0.
Thus, the straight line gives by (5),(1), and (3) are the radical axsed of the circles `S_(1)=0,S_(2)=0` and `S_(3)=0` taken in pairs. Since the radical axes of the three circles taken in pairs concurrent or parrallel, we have
`|{:(a-a',b-b',c-c'),(A,B,C),(A',B',C'):}|=0`