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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2003

  • question_answer
    If the area of the circle \[4{{x}^{2}}+4{{y}^{2}}-8x+16y+k=0\]is\[9\pi \]sq unit, then the value of k is:

    A)  4 sq unit       

    B)         16 sq unit           

    C)         \[-16\]sq unit     

    D)         \[\pm 16\] sq unit

    E)  none of these

    Correct Answer: C

    Solution :

    The equation of circle is \[{{x}^{2}}+{{y}^{2}}-2x+4y+\frac{k}{4}=0\] \[\therefore \]Radius of circle \[=\sqrt{1-4-\frac{k}{4}}\] Area of circle \[=9\pi \]                   (given) \[\Rightarrow \]               \[\pi \left( 5-\frac{k}{4} \right)=9\pi \] \[\Rightarrow \]               \[5-9=\frac{k}{4}\] \[\Rightarrow \]               \[k=-16\]


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