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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