Conic Sections Question 234
Question: Distances form the origin to the centres of the three circles $ x^{2}+y^{2}-2{\lambda _{i}}x=c^{2} $ (where c is constant and i= 1, 2, 3) are in GP. Then the lengths of tangents drawn from any point on the circle $ x^{2}+y^{2}=c^{2} $ to these circles are in
Options:
A) A.P.
B) GP.
C) H.P.
D) None
Show Answer
Answer:
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 $ (ccos\theta ,csin\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.
BETA
