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

  • question_answer
    The point at which the normal to the circle \[{{x}^{2}}+{{y}^{2}}+4x+6y-39=0\]at the point (2, 3) will meet the circle again, is

    A)            (6, -9)                                       

    B)            (6, 9)

    C)            (-6, -9)                                      

    D)            (-6, 9)

    Correct Answer: C

    Solution :

               Equation of normal will be \[\frac{x-2}{2+2}=\frac{y-3}{3+3}\]                    \[\Rightarrow \]\[3x-2y=0\Rightarrow x=\frac{2y}{3}\]                    Thus \[{{\left( \frac{2y}{3} \right)}^{2}}+{{y}^{2}}+4\left( \frac{2y}{3} \right)+6y-39=0\]                    \[\Rightarrow y=3,\ -9\Rightarrow x=2,\ -6\]                    Hence another point will be \[(-6,\,-9)\].


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