JEE Main & Advanced Mathematics Circle and System of Circles Question Bank Tangent and normal to a circle

  • question_answer
    The equation of pair of tangents to the circle \[{{x}^{2}}+{{y}^{2}}-2x+4y+3=0\]from \[(6,-5)\], is               [AMU 1980]

    A)            \[7{{x}^{2}}+23{{y}^{2}}+30xy+66x+50y-73=0\]

    B)            \[7{{x}^{2}}+23{{y}^{2}}+30xy-66x-50y-73=0\]

    C)            \[7{{x}^{2}}+23{{y}^{2}}-30xy-66x-50y+73=0\]

    D)            None of these

    Correct Answer: A

    Solution :

               \[S{{S}_{1}}={{T}^{2}}\]                    \[\Rightarrow ({{x}^{2}}+{{y}^{2}}-2x+4y+3)(36+25-12x-20y+3)\]                    \[={{(6x-5y-x-6+2(y-5)+3)}^{2}}\]                    \[\Rightarrow 7{{x}^{2}}+23{{y}^{2}}+30xy+66x+50y-73=0\].


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