A) \[ff=gg\]
B) \[fg=fg\]
C) \[{{(fg)}^{2}}={{(fg)}^{2}}\]
D) \[fg=fg\]
Correct Answer: D
Solution :
As both the circles pass through the origin and so they must have the same tangent at (0, 0). The general equation of tangent of the given circles are \[x{{x}_{1}}+y{{y}_{1}}+g\,(x+{{x}_{1}})+f\,(y+{{y}_{1}})\,=0\] \[x{{x}_{1}}+y{{y}_{1}}+g\,(x+{{x}_{1}})+f\,(y+{{y}_{1}})\,=0\] Substituting \[{{x}_{1}}=0\] and \[{{y}_{1}}=0\] we get \[gx+fy=0\Rightarrow gx+fy=0\] or \[\frac{f}{g}=\frac{f}{g}\] or \[gf=gf\]
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