JEE Main & Advanced JEE Main Paper (Held On 10 April 2015)

  • question_answer
    If \[y+3x=0\] is the equation of a chord of the circle, \[{{x}^{2}}+{{y}^{2}}-30x=0,\]then the equation of the circle with this chord as diameter is: JEE Main Online Paper (Held On 10 April 2015)

    A) \[{{x}^{2}}+{{y}^{2}}+3x+9y=0\]

    B) \[{{x}^{2}}+{{y}^{2}}-3x+9y=0\]

    C) \[{{x}^{2}}+{{y}^{2}}-3x-9y=0\]

    D) \[{{x}^{2}}+{{y}^{2}}+3x-9y=0\]

    Correct Answer: B

    Solution :

                     Given that \[y+3x=0\]equation of a choral of the circle then \[y=-3x\]                   ?(i) \[{{x}^{2}}+{{\left( -3x \right)}^{2}}-30x=0\] \[10{{x}^{2}}-30x=0\] \[10x\left( x-3 \right)=0\] \[x=0y=0\]so the equation of the circle is \[\left( x-3 \right)\left( x-0 \right)+\left( y+9 \right)\left( y-0 \right)=0\] \[{{x}^{2}}-3x+{{y}^{2}}+9=0\] \[{{x}^{2}}+{{y}^{2}}-3x+9y=0\]


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