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

  • question_answer
    The normal to the circle \[{{x}^{2}}+{{y}^{2}}-3x-6y-10=0\]at the point (?3, 4), is                                           [RPET 1986, 89]

    A)            \[2x+9y-30=0\]                      

    B)            \[9x-2y+35=0\]

    C)            \[2x-9y+30=0\]                      

    D)            \[2x-9y-30=0\]

    Correct Answer: A

    Solution :

               \[\frac{x-{{x}_{1}}}{{{x}_{1}}+g}=\frac{y-{{y}_{1}}}{{{y}_{1}}+f}\]\[\Rightarrow \frac{x+3}{-3-\frac{3}{2}}=\frac{y-4}{4-3}\Rightarrow 2x+9y-30=0\].


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