A) \[\frac{1}{a}+\frac{1}{b}=\frac{1}{c}\]
B) \[\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}=\frac{1}{{{c}^{2}}}\]
C) \[\frac{1}{a}+\frac{1}{b}={{c}^{2}}\]
D) \[\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}=\frac{1}{c}\]
Correct Answer: D
Solution :
\[{{C}_{1}}(-a,\ 0);\]\[{{C}_{2}}(0,\ -b);\] \[{{R}_{1}}(\sqrt{{{a}^{2}}-c});\] \[{{R}_{2}}(\sqrt{{{b}^{2}}-c})\] \[{{C}_{1}}{{C}_{2}}=\sqrt{{{a}^{2}}+{{b}^{2}}}\] Since they touch each other, therefore \[\sqrt{{{a}^{2}}-c}+\sqrt{{{b}^{2}}-c}=\sqrt{{{a}^{2}}+{{b}^{2}}}\] \[\Rightarrow {{a}^{2}}{{b}^{2}}-{{b}^{2}}c-{{a}^{2}}c\]= 0 Multiply by \[\frac{1}{{{a}^{2}}{{b}^{2}}c},\ \]we get \[\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}=\frac{1}{c}\].
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