Fitting circle in $f(x)=x^3-3x^2$ what is the radius? 
The whole problem is the above.
I tried to solve it analytically but I couldnot find enough equation to solve it.
Assume the center is $(a,-r)$ where $r$ is the radius of the circle.
The points on the sphere $(x-a)^2+(y+r)^2=r^2$
and $(x,f(x))$ must satisfy it so if we denote the intersection points $(x_1,y_1),(x_2,y_2)$:
$$(x_1-a)^2+(x_1^3-3x_1^2+r)^2=r^2$$
$$(x_2-a)^2+(x_2^3-3x_2^2+r)^2=r^2$$
Are two equation with 4 unknown.
Can you give me elegant relations between these two unknowns?


In figure suppose AC is straight line.We first find the angle BAC the curve makes with x axis:
$y=x^3-3x^2$ ⇒ $y'=3x^2-6x$
⇒ $tan (\widehat{BAC})=y'$ at point A(3, 0):
$tan (\widehat{BAC})=3^3-6\times 3=9$
⇒ $ \widehat{BAC} ≈83^o$
⇒$ \widehat{BAC}=\widehat BAD ≈41.5^o$
So the center of circle is on line $y=(m=tan 41.5)(x-3)$
Now we use plain geometry. we have:
$AB^2+(BD=r)^2=AD^2$
$BD^2+r^2=OD^2$
$AB+BO=3$
$(AB+BO)^2=AB^2+OB^2+2\cdot AB\cdot BO$
⇒ $AB^2+OB^2-2r^2=9-(DA^2+OD^2)$
$tan(41.5)=\frac r{AB}$
$OD^2=OB^2+r^2=(3-AB)^2+r^2$
These two relations give $OD^2$ in terms of r.
$sin(41.5)=\frac r{AD}$
This relation gives AD in term of r.
Now in relation:
$AB^2+OB^2-2r^2=9-(DA^2+OD^2)$
$OB=3-AB$
So OB can also be found in term of r. Plugging, DA, OA, AB and OB in above relation give an equation in terms of r
Note, r ≈1.1 and coordinates of its center is D( ≈1.8,  ≈-1.1)
