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

  • question_answer
    If the centre of a circle is (?6, 8) and it passes through the origin, then equation to its tangent at the origin, is [MNR 1976]

    A)            \[2y=x\]                                    

    B)            \[4y=3x\]

    C)            \[3y=4x\]                                 

    D)            \[3x+4y=0\]

    Correct Answer: B

    Solution :

               Centre (?6, 8), radius \[=\sqrt{{{6}^{2}}+{{8}^{2}}}=10\]                    Equation of circle \[{{x}^{2}}+{{y}^{2}}+12x-16y=0\]                    Equation of tangent at (0, 0) is\[3x-4y=0\] .


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