A) \[\frac{1}{{{b}^{2}}}+\frac{1}{{{c}^{2}}}=\frac{1}{{{a}^{2}}}\]
B) \[\frac{1}{{{c}^{2}}}+\frac{1}{{{a}^{2}}}=\frac{1}{{{b}^{2}}}\]
C) \[\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}=\frac{1}{{{c}^{2}}}\]
D) None of these
Correct Answer: C
Solution :
The two circles are \[{{x}^{2}}+{{y}^{2}}-2ax+{{c}^{2}}=0\] and \[{{x}^{2}}+{{y}^{2}}-2by+{{c}^{2}}=0\] centres:\[{{C}_{1}}(a,\,\,0)\] \[{{C}_{2}}(0,\,\,b)\] radii:\[{{r}_{1}}=\sqrt{{{a}^{2}}-{{c}^{2}}}\], \[{{r}_{2}}=\sqrt{{{b}^{2}}-{{c}^{2}}}\] Since, the two circles touch each other externally, therefore \[{{C}_{1}}{{C}_{2}}={{r}_{1}}+{{r}_{2}}\] \[\Rightarrow \] \[\sqrt{{{a}^{2}}+{{b}^{2}}}=\sqrt{{{a}^{2}}-{{c}^{2}}}+\sqrt{{{b}^{2}}-{{c}^{2}}}\] \[\Rightarrow \] \[{{a}^{2}}+{{b}^{2}}={{a}^{2}}-{{c}^{2}}+{{b}^{2}}-{{c}^{2}}\] \[\Rightarrow \] \[+2\sqrt{{{a}^{2}}-{{c}^{2}}}\sqrt{{{b}^{2}}-{{c}^{2}}}\] \[\Rightarrow \] \[{{c}^{4}}={{a}^{2}}{{b}^{2}}-{{c}^{2}}({{a}^{2}}+{{b}^{2}})+{{c}^{4}}\] \[\Rightarrow \] \[{{a}^{2}}{{b}^{2}}={{c}^{2}}({{a}^{2}}+{{b}^{2}})\] \[\Rightarrow \] \[\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}=\frac{1}{{{c}^{2}}}\]
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