Vertices of an isosceles triangle of area  $a^2$

and $(-a,0)$ and $(a,0).$  Equation of the circumcircle of the triangle is

detailed solution

Correct option is C

Since the triangle is isosceles, third vertex lies on the line x = 0, perpendicular to the base and passing through the mid-point (0, 0) of the base. As the area is a 2 , distance of the vertex from the base is a as the length of the base is 2a. So vertex of the triangle is (0, ± a) and let the equation of the circle passing through the vertices of the  triangle  x2+y2+2gx+2fy+c=0,  then a2+2ga+c=0and  a2±2fa+c=0  ⇒ c=−a2,g=f=0 and the equation of the required circle isx2+y2=a2