question_answer

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]

A)            \[f'g=g'f\]                                   

B)            \[fg=f'g'\]

C)            \[f'g'+fg=0\]                             

D)            \[f'g+g'f=0\]

Correct Answer: A

Solution :

           According to the figure, \[OA+{O}'A=OO'\]                    \[\sqrt{{{g}^{2}}+{{f}^{2}}}+\sqrt{f{{'}^{2}}+g{{'}^{2}}}=\sqrt{{{(g'-g)}^{2}}+{{(f'-f)}^{2}}}\]                    \[\Rightarrow {{g}^{2}}+{{f}^{2}}+f{{'}^{2}}+g{{'}^{2}}+2\sqrt{{{g}^{2}}+{{f}^{2}}}\times \sqrt{f{{'}^{2}}+g{{'}^{2}}}\]                    \[={{(g'-g)}^{2}}+{{(f'-f)}^{2}}\]                     \[\Rightarrow 2\sqrt{{{g}^{2}}+{{f}^{2}}}\sqrt{f{{'}^{2}}+g{{'}^{2}}}=-2(gg'+ff')\]                    \[\Rightarrow {{g}^{2}}f{{'}^{2}}+{{f}^{2}}g{{'}^{2}}=2gg'ff'\]. \[\therefore {{(gf'-fg')}^{2}}=0\Rightarrow gf'=fg'\].