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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 24

  • question_answer
    A circle S touches the line \[x+y=2\] at \[(1,1)\] and cuts the circle \[{{x}^{2}}+{{y}^{2}}+4x+5y-6=0\]P and Q, respectively. Then PQ always passes through the point

    A) \[(2,3)\]                 

    B)        \[(6,-4)\]     

    C) \[(-6,-4)\]              

    D)       \[(-6,4)\]

    Correct Answer: B

    Solution :

    [b] Equation of family of circles touching line \[x+y=2\] at point \[(1,1)\] is: \[{{(x-1)}^{2}}+{{(y-1)}^{2}}+\lambda (x+y-2)=0\] Equation of common chord PQ with the circle \[{{x}^{2}}+{{y}^{2}}+4x+5y-6=0\] is \[(-6x-7y+8)+\lambda (x+y-2)=0,\] which passes through the point \[(6,-4)\].         


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