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 :

Step-by-step solution

Let C =(x, y) Now, CA ^2= CB ^2= AB ^2 aligned & (x+a)^2+y^2=(x-a)^2+y^2=(2 a)^2 & x^2+2 a x+a^2+y^2=4 a^2 & and x^2-2 a x+a^2+y^2=4 a^2 aligned From (i) and (ii), x=0 and y= 3 a Since point C (x, y) lies above the x -axis and a>0 , hence y= 3 a C =(0, 3 a) Let the equation of circumcircle be x^2+y^2+2 g x+2 f y+ C =0 Since points A( -a, 0), B (a, 0) and C (0, 3 a) lie on the circle, therefore aligned & a^2-2 g a+ C =0 & a^2+2 g a+ C =0 aligned and 3 a^2+2 3 a f+ C =0 From (iii), (iv), and (v) g=0, c=-a^2, f=- a 3