JEE Main & Advanced Mathematics Pair of Straight Lines Question Bank Critical Thinking

  • question_answer The lines joining the origin to the points of intersection of the line \[y=mx+c\]and the circle \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\]will be mutually perpendicular, if                                             [Roorkee 1977]

    A)            \[{{a}^{2}}({{m}^{2}}+1)={{c}^{2}}\]                              

    B)            \[{{a}^{2}}({{m}^{2}}-1)={{c}^{2}}\]

    C)            \[{{a}^{2}}({{m}^{2}}+1)={{c}^{2}}\]                              

    D)            \[{{a}^{2}}({{m}^{2}}-1)=2{{c}^{2}}\]

    Correct Answer: C

    Solution :

               Making the equation of circle homogeneous with the help of line \[y=mx+c,\] we get \[{{x}^{2}}+{{y}^{2}}-{{a}^{2}}{{\left( \frac{y-mx}{c} \right)}^{2}}=0\].            \[\Rightarrow {{c}^{2}}{{x}^{2}}+{{c}^{2}}{{y}^{2}}-{{a}^{2}}{{y}^{2}}-{{a}^{2}}{{m}^{2}}{{x}^{2}}+2{{a}^{2}}mxy=0\]            \[\,\Rightarrow ({{c}^{2}}-{{a}^{2}}{{m}^{2}}){{x}^{2}}+({{c}^{2}}-{{a}^{2}}){{y}^{2}}-2{{a}^{2}}mxy=0\]       .....(i)            Hence lines represented by (i) are perpendicular, if            \[{{c}^{2}}-{{a}^{2}}{{m}^{2}}+{{c}^{2}}-{{a}^{2}}=0\,\Rightarrow \,\,2{{c}^{2}}={{a}^{2}}(1+{{m}^{2}})\].


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