If two vertices of an equilateral triangle are A(-a, 0) and B(a, 0), a > 0, and the third vertex C lies above x-axis, then the equation of the circumcircle of $∆$ ABC is

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$x^{2} + y^{2} - 2 a y = a^{2}$

$x^{2} + y^{2} - \sqrt{3} a y = a^{2}$

$3 x^{2} + 3 y^{2} - 2 \sqrt{3 a} y = 3 a^{2}$

$3 x^{2} + 3 y^{2} - 2 a y = 3 a^{2}$

answer is A.

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Detailed Solution

$L e t C (x, y) b e t h e t h i r d v e r t e x o f t r i a n g l e A B C h a v i n g t w o v e r t i c e s a t A (- a, 0) a n d B (a, 0) . S i n c e △ A B C i s e q u i l a t e r a l . T h e r e f o r e, A C = B C = A B N o w, A C = B C \Rightarrow {(x + a)}^{2} + {(y - 0)}^{2} = {(x - a)}^{2} + {(y - 0)}^{2} \Rightarrow {(x + a)}^{2} + y^{2} = {(x - a)}^{2} + y^{2} \Rightarrow 4 a x = 0 \Rightarrow x = 0 A g a i n, A C = B C = A B \Rightarrow A C = A B \Rightarrow {(x + a)}^{2} + {(y - 0)}^{2} = {(a + a)}^{2} + 0^{2} \Rightarrow {(x + a)}^{2} + y^{2} = 4 a^{2} \Rightarrow {(0 + a)}^{2} + y^{2} = 4 a^{2} \Rightarrow y^{2} = 3 a^{2} \Rightarrow y = \pm \sqrt{3} a H e n c e, t h e c o o r d i n a t e s o f t h e t h i r d v e r t e x a r e C (0, \sqrt{3} a ) (∵ a > 0 a n d t h i r d v e r t e x l i e s a b o v e x - a x i s) L e t t h e e q u a t i o n o f c i r c l e b e x^{2} + y^{2} + 2 g x + 2 f y + c = 0 \to (1) s u b A, B a n d C p o i n t s i n c i r c l e t h e n w e g e t g = 0, c = - a^{2}, f = - \frac{a}{\sqrt{3}} s u b t h e s e v a l u e s i n (1) t h e n w e g e t 3 x^{2} + 3 y^{2} - 2 \sqrt{3} a y = 3 a^{2}$