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

  • question_answer
    The gradient of the normal at the point (-2, -3) on the circle \[{{x}^{2}}+{{y}^{2}}+2x+4y+3=0\]is

    A)            1    

    B)            -1

    C)            \[\frac{3}{2}\]                          

    D)            \[\frac{1}{2}\]

    Correct Answer: A

    Solution :

               The equation of tangent at point \[(-2,\ -3)\] to the circle \[{{x}^{2}}+{{y}^{2}}+2x+4y+3=0\] is,                         \[-2x-3y+1(x-2)+2(y-3)+3=0\]                    \[\Rightarrow -2x-3y+x-2+2y-6+3=0\]                    \[\Rightarrow -x-y-5=0\Rightarrow x+y+5=0\]                    or\[y=-x-5\]; so, \[m=-1\]                    Hence, gradient of normal\[=\frac{-1}{-1}=1\].


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