If two vertices of an equilateral triangle are A(−a,0) and B(a,0), a>0, and the third vertex C lies above the x-axis, then the equation of the circumcircle of △ABC is:
Solutions · 1
Official solution
Since AB=2a, the third vertex is C=(0,3a) (above the axis). Let the circumcircle be x2+y2+2gx+2fy+c=0. Using A,B,C:
\[a^2-2ga+c=0,\quad a^2+2ga+c=0,\quad 3a^2+2\sqrt{3}af+c=0.\]
The first two give g=0,c=−a2. The third then gives f=−3a.
Circle: x2+y2−32ay−a2=0⇒3x2+3y2−23ay−3a2=0⇒3x2+3y2−23ay=3a2.