Circle And System Of Circles Question 240
Question: Circles $ x^{2}+y^{2}+2gx+2fy=0 $ and $ x^{2}+y^{2} $ $ +2g’x+2f’y= $ $ 0 $ touch externally, if
[MP PET 1994; Karnataka CET 2003]
Options:
A) $ f’g=g’f $
B) $ fg=f’g’ $
C) $ f’g’+fg=0 $
D) $ f’g+g’f=0 $
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Answer:
Correct Answer: A
Solution:
According to the figure, $ OA+{O}‘A=OO’ $
$ \sqrt{g^{2}+f^{2}}+\sqrt{f{{’}^{2}}+g{{’}^{2}}}=\sqrt{{{(g’-g)}^{2}}+{{(f’-f)}^{2}}} $
$ \Rightarrow g^{2}+f^{2}+f{{’}^{2}}+g{{’}^{2}}+2\sqrt{g^{2}+f^{2}}\times \sqrt{f{{’}^{2}}+g{{’}^{2}}} $
$ ={{(g’-g)}^{2}}+{{(f’-f)}^{2}} $
$ \Rightarrow 2\sqrt{g^{2}+f^{2}}\sqrt{f{{’}^{2}}+g{{’}^{2}}}=-2(gg’+ff’) $
$ \Rightarrow g^{2}f{{’}^{2}}+f^{2}g{{’}^{2}}=2gg’ff’ $ .
$ \therefore {{(gf’-fg’)}^{2}}=0\Rightarrow gf’=fg’ $ .
BETA
