A) \[\frac{1}{a}+\frac{1}{b}=\frac{1}{c}\]
B) \[\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}=\frac{1}{c}\]
C) \[\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}=\frac{1}{{{c}^{2}}}\]
D) \[\frac{1}{{{a}^{2}}}-\frac{1}{{{b}^{2}}}-\frac{1}{c}\]
E) \[\frac{1}{a}+\frac{1}{b}=\frac{1}{{{c}^{2}}}\]
Correct Answer: B
Solution :
Centre and radius of circle\[{{x}^{2}}+{{y}^{2}}+2ax+c=0\]are\[(-a,0)\]and\[\sqrt{{{a}^{2}}-c}\]respectively and centre and radius of another circle\[{{x}^{2}}+{{y}^{2}}+2by+c=0\]are\[(0,-b)\]and\[\sqrt{{{b}^{2}}-c}\]respectively. If both the circles touches each other, then \[\sqrt{{{a}^{2}}+{{b}^{2}}}=\sqrt{{{a}^{2}}-c}+\sqrt{{{b}^{2}}-c}\] \[\Rightarrow \] \[{{c}^{2}}=({{a}^{2}}-c)({{b}^{2}}-c)\] \[\Rightarrow \] \[{{a}^{2}}c+{{b}^{2}}c={{a}^{2}}{{b}^{2}}\] \[\Rightarrow \] \[\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}=\frac{1}{c}\]
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