RAJASTHAN ­ PET Rajasthan PET Solved Paper-2003

  • question_answer
    Intersection points of the circles\[{{x}^{2}}+{{y}^{2}}=25\] and \[{{x}^{2}}+{{y}^{2}}-8x+7=0,\]are

    A)  (4, 3) and \[(4,-3)\]

    B) \[(4,-3)\]and\[(-4,-3)\]

    C) \[(-4,3)\]and (4, 3)

    D)  (4, 3) and (3, 4)

    Correct Answer: A

    Solution :

     Given,         \[{{x}^{2}}+{{y}^{2}}=25\]           ...(i) and             \[{{x}^{2}}+{{y}^{2}}-8x+7=0\]              ...(ii) From Eqs. (i) and (ii), \[25-8x+7=0\] \[\Rightarrow \] \[8x=32\] \[\Rightarrow \]    \[x=4\]and\[y=\pm 3\] Hence, intersection points are (4,3) and\[(4,-3)\].


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