If the normals to the curve $y=x^2$ at the points $P,Q$ and $R$ pass through the point $(0,\frac{3}{2}),$ and the equation of the circle circumscribing the triangle $PQR$ is $x^2+y^2+2gx+2fy+c=0,$ then find the value of  $(g^2+f^2+c^2).$

 $y=x^2;x=t;y=t^2$

 $\frac{dy}{dx}=2x=2t$
$\therefore$  slope of normal,  $m=\frac{-1}{2t}$
Equation of normal  $y-t^2=\frac{1}{2t}(x-t)$
Or       $2t(y-t^2)=-x+t$
If         $x=0;y=\frac{3}{2}$
 $2t(\frac{3}{2}-t^2)=t\Rightarrow t=0$
Or       $3-2t^2=1\Rightarrow at=0$
Hence, one of the point is origin and the other two are (-1, 1) and (1, 1)$\Rightarrow PQR$
 is a right triangle.
  $\therefore$   Radius of the circle is 1.
Its equation is  $x^2+{(y-1)}^2=1\Rightarrow x^2+y^2-2y=0$