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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 15

  • question_answer
    Let \[P({{x}_{1}},{{y}_{1}})\]and \[Q({{x}_{2}},{{y}_{2}})\]are two points such that their abscissa \[{{x}_{1}}\]and \[{{x}_{2}}\]are the roots of the equation \[{{x}^{2}}+2x-3=0\]while the ordinates \[{{y}_{1}}\] and \[{{y}_{2}}\]are the roots of the equation \[{{y}^{2}}+4y-12=0\]. The centre of the circle with PQ as diameter is

    A) (-1,-2)             

    B) (1,2)

    C)  (1,-2)             

    D)  (-1,2)

    Correct Answer: A

    Solution :

    : Since,\[{{x}_{1}},{{x}_{2}}\]are roots of \[{{x}^{2}}+2x-3=0\] \[\therefore \]\[{{x}_{1}}+{{x}_{2}}=-2\Rightarrow \frac{{{x}_{1}}+{{x}_{2}}}{2}=1\] Also, \[{{y}_{1}},{{y}_{2}}\]are roots of \[{{y}^{2}}+4y-12=0\] \[\therefore \]\[{{y}_{1}}+{{y}_{2}}=-4\Rightarrow \frac{{{y}_{1}}+{{y}_{2}}}{2}=-2\] Centre of circle\[\left( \frac{{{x}_{1}}+{{x}_{2}}}{2},\frac{{{y}_{1}}+{{y}_{2}}}{2} \right)=(-1,-2)\]


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