JEE Main & Advanced JEE Main Paper (Held On 22 April 2013)

  • question_answer
    If two vertices of an equilateral triangle are \[\operatorname{A}(-a,0)\] and \[B(a,\,0),\,a>0\], the third vertex C lies above \[x-\]axis then the equation of the circumcircle of   \[\Delta \operatorname{ABC}\]       is :       JEE Main  Online Paper (Held On 22 April 2013)

    A)  \[3{{x}^{2}}+3{{y}^{2}}-2\sqrt{3}ay=3{{a}^{2}}\]

    B)  \[3{{x}^{2}}+3{{y}^{2}}-2ay=3{{a}^{2}}\]

    C)  \[{{x}^{2}}+{{y}^{2}}-2ay={{a}^{2}}\]

    D)  \[{{x}^{2}}+{{y}^{2}}-\sqrt{3}ay={{a}^{2}}\]

    Correct Answer: A

    Solution :

     Let \[C=(x,y)\] Now, \[C{{A}^{2}}=C{{B}^{2}}=A{{B}^{2}}\] \[\Rightarrow \] \[{{(x+a)}^{2}}+{{y}^{2}}={{(x-a)}^{2}}+{{y}^{2}}={{(2a)}^{2}}\] \[\Rightarrow \]\[{{x}^{2}}+2ax+{{a}^{2}}+{{y}^{2}}=4{{a}^{2}}\]                               ?(i) and \[{{x}^{2}}-2ax+{{a}^{2}}+{{y}^{2}}=4{{a}^{2}}\]                        ?(ii) From (i) and (ii), \[x=0\]and \[y=\pm \sqrt{3}a\] Since point \[C(x,y)\]lies above the \[x-\]axis and \[a>0,\]hence \[y=\sqrt{3}a\] \[\therefore \] \[C=(0,\sqrt{3}a)\] Let the equation of circumcircle be \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+C=0\] Since points \[A(-a,0),B(a,0)\]and \[C(0,\sqrt{3a})\]lie on the circle, therefore \[{{a}^{2}}-2ga+C=0\]                    ?(iii) \[{{a}^{2}}+2ga+C=0\]                   ?(iv) and \[3{{a}^{2}}+2\sqrt{3}af+=0\]                            ?(v) From (iii),(iv), and (v) \[g=0,c=-{{a}^{2}},f=-\frac{a}{\sqrt{3}}\] Hence equation of the circumcircle is \[{{x}^{2}}+{{y}^{2}}-\frac{2a}{\sqrt{3}}y-{{a}^{2}}=0\] \[\Rightarrow \]\[{{x}^{2}}+{{y}^{2}}-\frac{2\sqrt{3}ay}{3}-{{a}^{2}}=0\] \[\Rightarrow \]\[3{{x}^{2}}+3{{y}^{2}}-2\sqrt{3}ay=3{{a}^{2}}\]

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