JEE Main & Advanced AIEEE Solved Paper-2004

  • question_answer
    If the lines\[2x+3y+1=0\]and\[3x-y-4=0\]lie along diameters of a circle of circumference\[10\pi ,\]then the equation of the circle is

    A) \[{{x}^{2}}+{{y}^{2}}-2x+2y-23=0\]

    B) \[{{x}^{2}}+{{y}^{2}}-2x-2y-23=0\]

    C) \[{{x}^{2}}+{{y}^{2}}+2x+2y-23=0\]

    D) \[{{x}^{2}}+{{y}^{2}}+2x-2y-23=0\]

    Correct Answer: A

    Solution :

    The lines\[2x+3y+1=0\]and\[3x-y-4=0\] are diameters of circle. On solving these equations, we get \[x=1\]and \[y=-1\] Therefore, the centre of circle is\[(1,-1)\] and circumference of circle\[=10\pi .\] \[\Rightarrow \]               \[2\pi r=10\pi \] \[\Rightarrow \]               \[r=5\] \[\therefore \]Equation of circle is \[{{(x-{{x}_{1}})}^{2}}+{{(y-{{y}_{1}})}^{2}}={{r}^{2}}\] \[\Rightarrow \]               \[{{(x-1)}^{2}}+{{(y-1)}^{2}}={{5}^{2}}\] [\[\because \]\[r=5,\].centre\[(1,-1)\]] \[\Rightarrow \] \[{{x}^{2}}+1-2x+{{y}^{2}}+2y+1=25\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}-2x+2y-23=0\]

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