• # 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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