• # 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)  ${{x}^{2}}+{{y}^{2}}-\sqrt{3}ay={{a}^{2}}$

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}}$