question_answer

The two circles which passes through \[(0,a)\]and \[(0,-a)\]and touch the line \[y=mx+c\] will intersect each other at right angle, if

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\[{{c}^{2}}={{a}^{2}}(2+{{m}^{2}})\].