JEE Main & Advanced JEE Main Paper (Held On 19 April 2014)

  • question_answer
    The equation of circle described on the chord \[3x+y+5=0\]of the circle \[{{x}^{2}}+{{y}^{2}}=16\]as diameter is:     JEE Main Online Paper (Held On 19 April 2016)

    A) \[{{x}^{2}}+{{y}^{2}}+3x+y-11=0\]

    B) \[{{x}^{2}}+{{y}^{2}}+3x+y+1=0\]

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

    D) \[{{x}^{2}}+{{y}^{2}}+3x+y-22=0\]

    Correct Answer: A

    Solution :

    Given circle is \[{{x}^{2}}+{{y}^{2}}-16=0\] Eqn of chord say AB of given circle is \[3x+y+5=0.\] Equation of required circle is \[{{x}^{2}}+{{y}^{2}}-16+\lambda (3x+y+5)=0\] \[\Rightarrow \]\[{{x}^{2}}+{{y}^{2}}+(3\lambda )x+(\lambda )y+5\lambda -16=0\]?(1) Centre \[C=\left( \frac{-3\lambda }{2},\frac{-\lambda }{2} \right).\] If line AB is the diameter of circle (1), then \[C=\left( \frac{-3\lambda }{2},\frac{-\lambda }{2} \right)\]will lie on line AB. i.e. \[3\left( \frac{-3\lambda }{2} \right)+\left( \frac{-\lambda }{2} \right)+5=0\] \[\Rightarrow \]\[\frac{9\lambda -\lambda }{2}+5=0\Rightarrow \lambda =1\] Hence, required eqn of circle is \[{{x}^{2}}+{{y}^{2}}+3x+y+5-16=0\] \[\Rightarrow \]\[{{x}^{2}}+{{y}^{2}}+3x+y-11=0\]

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