Let C=(x,y) Now,CA2=CB2=AB2 ⇒(x+a)2+y2=(x−a)2+y2=(2a)2 ⇒x2+2ax+a2+y2=4a2 and x2−2ax+a2+y2=4a2 From (i) and (ii), x=0 and y=±√3a Since point C(x,y) lies above the x -axis and a>0, hence y=√3a∴C=(0,√3a) Let the equation of circumcircle be x2+y2+2gx+2fy+C=0 Since points A(−a,0),B(a,0) and C(0,√3a) lie on the circle, therefore a2−2ga+C=0 a2+2ga+C=0 and 3a2+2√3af+C=0 From (iii), (iv), and (v) g=0,c=−a2,f=−