A) \[{{x}^{2}}+{{y}^{2}}-ax=0\]
B) \[{{x}^{2}}+{{y}^{2}}=\frac{1}{{{a}^{2}}}\]
C) \[{{y}^{2}}=4ax\]
D) \[{{x}^{2}}+{{y}^{2}}-ax-ay+{{a}^{2}}=0\]
E) \[{{x}^{2}}-{{y}^{2}}={{a}^{2}}\]
Correct Answer: B
Solution :
\[y=mx+c\]is tangent to\[{{x}^{2}}+{{y}^{2}}={{a}^{2}},\]if \[c=\pm a\sqrt{1+{{m}^{2}}}\] \[\therefore \]\[y=-\frac{lx}{m}+\frac{1}{m}\]is tangent to\[{{x}^{2}}+{{y}^{2}}={{a}^{2}},\]if \[\frac{1}{m}=\pm \frac{a}{m}\sqrt{{{l}^{2}}+{{m}^{2}}}\] \[\Rightarrow \] \[{{l}^{2}}+{{m}^{2}}=\frac{1}{{{a}^{2}}}\] \[\therefore \]Locus of\[(l,m)\]is\[{{x}^{2}}+{{y}^{2}}=\frac{1}{{{a}^{2}}}.\]
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