P,Q,R are the centres and r1,r2,r3 are the radii respectively of three coaxial circles, show that r21QR+r22RP+r23PQ=PQ.QR.RP.
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Solution
By special choice of axes the equation of coaxial system of circles can be put in the form x2+y2+2grx+c=0 where r=1,2,3. The co-ordinates of the centres of the three circles are P(−g1,0),Q(−g2,0),R(−g3,0) ∴QR=(−g3)−(−g2)=g2−g3 RP=g3−g1 and PQ=g1−g2. If the fixed point from where the tangents be drawn be (h,k), then t21=h2+k2+2g1h+c etc. ∴QRt21+RPt22+PQt23 =(g2−g3){h2+k2+c+2g1h}+....+....=0 or (h2+k2+c)∑(g2−g3)+2h∑g2(g1−g3)=0 ∵∑(g2−g3)=0 and ∑g1(g2−g3)=0.
From part (a), r21=(g1)2−c etc. ∴r21.QR+r22.RP+r23.PQ =(g21−c)(g2−g3)+(g22−c)(g3−g1)+(g23−c)(g1−g2) =∑g21(g2−g3)−c∑(g2−g3) =−(g1−g2)(g2−g3)(g3−g1)−c.0 =−PQ.QR.RP.