Mathematics - Circle Question with Solution | TestHub

The first circle has equation $(x-1{)}^2+(y-1{)}^2+\lambda (x+y-2)=0$. Its common chord with the second circle lies on $(\lambda -6)x+(\lambda -7)y+8-2\lambda =0$. This cannot be parallel to $x+y-2=0$. Perpendicularity requires $\lambda =13/2$, when the circles do not intersect and there is no chord $PQ$. The line passes through the simultaneous intersection of $x+y-2=0$ and $-6x-7y+8=0$, namely $(6,-4)$.