JEE Main & Advanced JEE Main Paper (Held On 22 April 2013)

  • question_answer
    If a circle C passing though (4, 0) touches the circle \[{{x}^{2}}+{{y}^{2}}+4x-6y-12=0\] externally at point (1, -1), then the radius of the circle C is:     JEE Main  Online Paper (Held On 22 April 2013)

    A) 5                                             

    B)  \[2\sqrt{5}\]

    C)   4                                           

    D)  \[\sqrt{57}\]

    Correct Answer: A

    Solution :

     Let A be the centre of given circle and B be the centre of circle C. \[{{x}^{2}}+{{y}^{2}}+4x-6y-12=0\] \[\therefore \] \[A=(-2,3)\]and \[B=(g,f)\] Now, from the figure, we have \[\frac{-2+g}{2}=1\]and \[\frac{3+f}{2}=-1\] (By mid point formula) \[\Rightarrow \]\[g=4\]and \[f=-5\] Now, required radius \[=OB=\sqrt{9+16}=\sqrt{25}=5\]

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