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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2003

  • question_answer
    If the circle\[{{x}^{2}}+{{y}^{2}}+2ax+8y+16=0\]touches the\[x-\]axis, then a is equal to

    A)  ±16             

    B)  ± 8

    C)  ± 1              

    D)  ± 4

    Correct Answer: D

    Solution :

     We know that, if the circle \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0,\]touches\[x-\]axis, then \[{{g}^{2}}=c\] Since,\[{{x}^{2}}+{{y}^{2}}+2ax+8y+16=0\]touches\[x-\]axis \[\therefore \] \[{{a}^{2}}=16\] \[\Rightarrow \] \[a=\pm 4\]


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