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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 19

  • question_answer
    Every tangent of a circle is perpendicular to exactly one member of the family of lines \[(x+y-2)+\lambda (7x-3y-4)=0\] at the point of contact of tangent. Also, the circle touches only one member of the family \[(2x-3y)+\mu (x-y-1)=0\]. The circle passes through the point

    A) \[(2,0)\]                 

    B)        \[(-1,0)\]              

    C) \[(0,2)\]                 

    D)        None of these

    Correct Answer: B

    Solution :

      [b] The given conditions in the question meet if first family of lines represents the diameters and fixed point of second family lies on the required circle. So, centre of the circle is \[C(1,1)\] and point on the circle is \[A(3,2)\]. Therefore, the equation of circle is: \[{{(x-1)}^{2}}+{{(y-1)}^{2}}=5\]


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