A) \[{{x}^{2}}+{{y}^{2}}=1\]
B) \[{{r}^{2}}({{x}^{2}}+{{y}^{2}})=1\]
C) \[{{x}^{2}}+{{y}^{2}}={{r}^{2}}\]
D) \[{{x}^{2}}+{{y}^{2}}=2{{r}^{2}}\]
Correct Answer: B
Solution :
The given line\[lx+my=1\Rightarrow y=-\frac{1}{m}x+\frac{1}{m}\]is the tangent to the circle\[{{x}^{2}}+{{y}^{2}}={{r}^{2}}\]. \[\therefore \] \[\frac{1}{m}=\pm r\sqrt{1+{{\left( -\frac{l}{m} \right)}^{2}}}\] \[[\because c=\pm \sqrt{1+{{m}^{2}}}]\] \[\Rightarrow \] \[\frac{1}{m}=\frac{\pm r\sqrt{{{m}^{2}}+{{l}^{2}}}}{m}\] \[\Rightarrow \] \[r\sqrt{{{m}^{2}}+{{l}^{2}}}=1\] \[\Rightarrow \] \[{{r}^{2}}({{l}^{2}}+{{m}^{2}})=1\] Hence, locus of\[(l,m)\]is \[{{r}^{2}}({{x}^{2}}+{{y}^{2}})=1\]
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