Polar of any point P w.r.t two given circles meet at Q. Prove that radical axis of circles bisect the line segment PQ

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You question seems incomplete. So Icannot give exact solution.Let equations of circle be:S1:x2+y2+2g1x+2f1y+c1=0S2:x2+y2+2g2x+2f2y+c2=0Let pole i.e. point Pbe x1,y1We know that equation of polar is:T=0For C1 we have:xx1+yy1+2g1x+x12+2f1y+y12+c1=0xx1+yy1+g1x+x1+f1y+y1+c1=0 iFor C1 we have:xx1+yy1+2g2x+x12+2f2y+y12+c2=0xx1+yy1+g2x+x1+f2y+y1+c2=0 iiEquate i and iixx1+yy1+g1x+x1+f1y+y1+c1=xx1+yy1+g2x+x1+f2y+y1+c2g1+f1y+y1+c1=g2x+x1+f2y+y1+c2x+x1g1-g2+y+y1f1-f2+c1-c2=0xg1-g2+x1g1-g2+yf1-f2+y1f1-f2+c1-c2=0xg1-g2+yf1-f2+x1g1-g2+y1f1-f2+c1-c2=02xg1-g2+2yf1-f2+2x1g1-g2+2y1f1-f2+2c1-2c2=0 ;iiiRadical axis of two circles will be:2g1-g2x+2f1-f2y+c1-c2=0 ;ivBy iii and iv, we get:c2-c1+2x1g1-g2+2y1f1-f2+2c1-2c2=0 2x1g1-g2+2y1f1-f2+c1-c2=0 x1g1-g2+y1f1-f2+c1-c22=0 

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