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Question

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)=g2g3
RP=g3g1 and PQ=g1g2.
If the fixed point from where the tangents be drawn be (h,k), then
t21=h2+k2+2g1h+c etc.
QRt21+RPt22+PQt23
=(g2g3){h2+k2+c+2g1h}+....+....=0
or (h2+k2+c)(g2g3)+2hg2(g1g3)=0
(g2g3)=0 and g1(g2g3)=0.

From part (a), r21=(g1)2c etc.
r21.QR+r22.RP+r23.PQ
=(g21c)(g2g3)+(g22c)(g3g1)+(g23c)(g1g2)
=g21(g2g3)c(g2g3)
=(g1g2)(g2g3)(g3g1)c.0
=PQ.QR.RP.

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