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

  • question_answer
    The equations of the tangents to the circle \[{{x}^{2}}+{{y}^{2}}-6x+4y=12\]which are parallel to the straight line \[4x+3y+5=0\], are   [ISM Dhanbad 1973; MP PET 1991]

    A)            \[3x-4y-19=0,\,\,3x-4y+31=0\]

    B)            \[4x+3y-19=0,\,\,4x+3y+31=0\]

    C)            \[4x+3y+19=0,\,\,4x+3y-31=0\]

    D)            \[3x-4y+19=0,3x-4y+31=0\]

    Correct Answer: C

    Solution :

               Let equation of tangent be\[4x+3y+k=0\], then \[\sqrt{9+4+12}=\left| \frac{4(3)+3(-2)+k}{\sqrt{16+9}} \right|\]                    \[\Rightarrow 6+k=\pm 25\Rightarrow k=19\]and\[-31\].                    Hence the tangents are \[4x+3y+19=0\] and \[4x+3y-31=0\].


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