JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Morning)

  • question_answer The line x = y touches a circle at the point (1, 1). If the circle also passes through the point (1, -3), then its radius is : [JEE Main 10-4-2019 Morning]

    A) \[3\sqrt{2}\]

    B) \[3\]

    C) \[2\sqrt{2}\]                

    D) \[2\]

    Correct Answer: A

    Solution :

    Equation of circle can be written as \[{{\left( x1 \right)}^{2}}+{{\left( y1 \right)}^{2}}+\lambda \left( xy \right)=0\] It passes through (1, -3) \[16+\lambda \left( 4 \right)=0\Rightarrow \lambda =-4\] So \[{{\left( x1 \right)}^{2}}+{{\left( y1 \right)}^{2}}4\left( xy \right)=0\] \[\Rightarrow {{x}^{2}}+{{y}^{2}}6x+2y+2=0\] \[\Rightarrow r=2\sqrt{2}\] (correct key is b)                 


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