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

  • question_answer
    Given the circles \[{{x}^{2}}+{{y}^{2}}-4x-5=0\]and\[{{x}^{2}}+{{y}^{2}}+6x-2y+6=0\]. Let P be a point \[(\alpha ,\beta )\]such that the tangents from P to both the circles are equal, then

    A)            \[2\alpha +10\beta +11=0\]

    B)            \[2\alpha -10\beta +11=0\]

    C)            \[10\alpha -2\beta +11=0\]  

    D)            \[10\alpha +2\beta +11=0\]

    Correct Answer: C

    Solution :

               Accordingly, \[{{\alpha }^{2}}+{{\beta }^{2}}-4\alpha -5={{\alpha }^{2}}+{{\beta }^{2}}+6\alpha -2\beta +6\]                    \[\Rightarrow 10\alpha -2\beta +11=0\].


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