A) A.P.
B) GP.
C) H.P.
D) None
Correct Answer: B
Solution :
[b] Centres of the circle are \[({{\lambda }_{1}},0),\,\,({{\lambda }_{2}},0)\] and \[({{\lambda }_{3}},0)\] \[\therefore \] Given \[{{\lambda }^{2}}_{2}={{\lambda }_{1}}{{\lambda }_{3}}\] Any point on \[{{x}^{2}}+{{y}^{2}}={{c}^{2}}\] is \[(c\,cos\theta ,c\,sin\theta )\] The length of tangents form this point to the given circles are: \[{{t}^{2}}_{1}=-2{{\lambda }_{1}}c\cos \theta ,{{t}^{2}}_{2}=2{{\lambda }_{2}}c\cos \theta ,\] \[{{t}^{2}}_{3}=-2{{\lambda }_{3}}c\cos \theta \] Clearly, \[{{(t_{2}^{2})}^{2}}=t_{1}^{2}\,.\,t_{3}^{2}\,\,\,\,\Rightarrow \,\,\,(\because \,\,\,\lambda _{2}^{2}\,\,=\,\,{{\lambda }_{1}}{{\lambda }_{3}})\] \[\therefore \,\,\,\,\,\,t_{1}^{2},\,\,t_{2}^{2},\,\,t_{3}^{2}\] are in G.P. hence \[\,{{t}_{1}},\,\,{{t}_{2}},\,\,{{t}_{3}}\] are also in GP.
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