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BCECE Engineering BCECE Engineering Solved Paper-2012

  • question_answer
    The other end of the diameter through the point \[(-1,1)\] on the circle \[{{x}^{2}}+{{y}^{2}}-6x+4y-12y=0\]is

    A) \[(-7,5)\]                             

    B)         \[(-7,-\text{ }5)\]                           

    C)        \[(7,-\text{ }5)\]              

    D)                         \[(7,\text{ }5)\]

    Correct Answer: C

    Solution :

    Given, circle is \[{{x}^{2}}+{{y}^{2}}-6x+4y-12=0\] Centre of this circle is \[(3,-2)\] Let other end of the diameter is \[(\alpha ,\beta ).\] \[\therefore \]  \[\frac{\alpha -1}{2}=3,\frac{\beta +1}{2}=-2\] \[\Rightarrow \]               \[\alpha =7,\beta =-5\] \[\therefore \]Other end of the diameter is \[(7,-5).\]


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