JEE Main & Advanced AIEEE Solved Paper-2006

  • question_answer
    If the lines\[3x-4y-7=0\]and\[2x-3y-5=0\]are two diameters of a circle of area\[49\,\pi \,sq\]units, the equation of the circle is       AIEEE  Solved  Paper-2006

    A) \[{{x}^{2}}+{{y}^{2}}+2x-2y-62=0\]

    B) \[{{x}^{2}}+{{y}^{2}}-2x+2y-62=0\]

    C) \[{{x}^{2}}+{{y}^{2}}-2x+2y-47=0\]

    D) \[{{x}^{2}}+{{y}^{2}}+2x-2y-47=0\]

    Correct Answer: C

    Solution :

    The intersection point of diameters of a circle is the centre of a circle. The given equations of diameters are \[3x-4y-7=0\]                                                   ...(i) and       \[2x-3y-5=0\]                                   ...(ii) On solving Eqs. (i) and (ii). we get \[x=1\]and\[y=-1\] \[\therefore \]Centre of circle is (1, -1). Let r be the radius of circle, then \[\pi {{r}^{2}}=49\pi \] \[\Rightarrow \] \[r=7\]units \[\therefore \]Equation of required circle is \[{{(x-1)}^{2}}+{{(y+1)}^{2}}=49\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}-2x+2y+1+1=49\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}-2x+2y-47=0\]

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