A) \[{{a}^{2}}={{c}^{2}}(2m+1)\]
B) \[{{a}^{2}}={{c}^{2}}(2+{{m}^{2}})\]
C) \[{{c}^{2}}={{a}^{2}}(2+{{m}^{2}})\]
D) \[{{c}^{2}}={{a}^{2}}(2m+1)\]
Correct Answer: C
Solution :
Equation of circles \[[{{x}^{2}}+(y-a)(y+a)]+\lambda x=0\Rightarrow {{x}^{2}}+{{y}^{2}}+\lambda x-{{a}^{2}}=0\] and \[\sqrt{{{\left( \frac{\lambda }{2} \right)}^{2}}+{{a}^{2}}}=\frac{\frac{-m\lambda }{2}+c}{\sqrt{1+{{m}^{2}}}}\] \[\Rightarrow (1+{{m}^{2}})\text{ }\left[ \frac{{{\lambda }^{2}}}{4}+{{a}^{2}} \right]={{\left( \frac{m\lambda }{2}-c \right)}^{2}}\] \[\Rightarrow (1+{{m}^{2}})\text{ }\left[ \frac{{{\lambda }^{2}}}{4}+{{a}^{2}} \right]=\frac{{{m}^{2}}{{\lambda }^{2}}}{4}-mc\lambda +{{c}^{2}}\] \[\Rightarrow {{\lambda }^{2}}+4mc\lambda +4{{a}^{2}}(1+{{m}^{2}})-4{{c}^{2}}=0\] \[\therefore \]\[{{\lambda }_{1}}{{\lambda }_{2}}=4[{{a}^{2}}(1+{{m}^{2}})-{{c}^{2}}]\Rightarrow {{g}_{1}}{{g}_{2}}=[{{a}^{2}}(1+{{m}^{2}})-{{c}^{2}}]\] and\[{{g}_{1}}{{g}_{2}}+{{f}_{1}}{{f}_{2}}=\frac{{{c}_{1}}+{{c}_{2}}}{2}\Rightarrow {{a}^{2}}(1+{{m}^{2}})-{{c}^{2}}=-{{a}^{2}}\] Hence\[{{c}^{2}}={{a}^{2}}(2+{{m}^{2}})\].You need to login to perform this action.
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