The minimum area of circle which touches the parabolas $y=x^2+1$  and $y^2=x-1$  is
 

Required area of circle is shortest distance between two curves as diameter of circle.Shortest distance is measured along the common normal of the curves.$y=x^2+1,y^2=x-1$  are symmetric about the line y = x.
 common Tangent to parabola and circle is parallel to y = x and line perpendicular to it is $y=-x+k$  
$\Rightarrow y=-(x-1)+k-1$, it will be normal to $y^2=x-1$ 
if  $k-1=-2am-a{m}^2=-2.\frac{1}{4}(-1)-\frac{1}{4}{(-1)}^3=\frac{3}{4}$
$\Rightarrow k=\frac{7}{4};$thus ,the normal is $y=-x+\frac{7}{4},$ on solving with the given parabolas we get $P=(\frac{1}{2},\frac{5}{4})andQ=(\frac{5}{4},\frac{1}{2})$


$PQ=\frac{3\sqrt{2}}{4}\Rightarrow$minimum radiusof circle is  $\frac{3\sqrt{2}}{8}$
Minimum area of the circle is  $\frac{9\pi }{32}$